Stoichiometry - AP Chemistry

Card 0 of 705

Question

Balance the following equation:

NaOH + H2SO4 → H2O + Na2SO4

Answer

Balancing equations:

there is 1 NaOH and 1 H2SO4 in the reactants; and 1 H2O and 1 Na2SO4 in the products

steps:

balance the Na first; get 2 NaOH

balance H next; get 1 H2SO4, 2 H2O

check to make sure that O and S balance

Compare your answer with the correct one above

Question

_Fe2O3 + _HCl ⇌ _FeCl3 + _H2O

The Following question will be based on the unbalanced reaction above

For the reaction above to be balanced what coefficient should be in front of the compound HCl?

Answer

Balancing reactions is best acheived by using a stepwise approach. It is useful to work through each atom making sure it is present in a balanced fashion on both sides of the equation. Starting with Fe, Fe2O3 is the only molecule with Fe present on the left side of the equation and FeCl3 is the only molecule with Fe present on the right side of the equation. Thus the molar ratio of Fe2O3 to FeCl3 must be 1:2. Moving on to O, the O on the left side of the equation exists as Fe2O3 and H2O on the right. The molar ratio of Fe2O3 to H2O is therefore 1:3. Cl exists in HCl on the left and FeCl3 on the right. Thus the molar ratio of HCl to FeCl3 must be 3:1. To ensure that these molar ratios are maintaned the balanced formula is then determined to be Fe2O3 + 6HCl -> 2FeCl3 + 3H2O

Compare your answer with the correct one above

Question

Balance the following equation:

FeCl3 + NOH5 → Fe(OH)3 + NH4Cl

Answer

The first thing to do is to balance the Cl (3 on the left; one on the right) ← add 3 for NH4Cl

Now, there are 3 N on the right and only one on the left; add a 3 to the NOH5

Check to see that the Fe, H, and O balance, which they do

Compare your answer with the correct one above

Question

After the following reaction is balanced, how many moles of H2O can 4 moles of C2H6 produce?

C2H6(s) + _O2(g) → _H2O(l) + _CO2(g)

Answer

The reaction balances out to 2C2H6(s) + 7O2(g) → 6H2O(l) + 4CO2(g). For every 2 moles of C2H6 you can produce 6 moles of H2O. Giving you a total production of 12 moles when you have 4 moles of C2H6.

Compare your answer with the correct one above

Question

Consider the following unbalanced equation for the combustion of propane, $C_3H_8$ :

$C_3H_{8\hspace{1 mm}(g)}+O_{2\hspace{1 mm}(g)}\rightarrow CO_{2\hspace{1 mm}(g)}+H_2O_{(l)}$

If you were to combust one mole of propane, how many moles of water would you produce?

Answer

Begin by balancing the equation. There are many ways to do this, but one method that is particularly useful is to assume that you have 1 mole of your hydrocarbon (propane), and balance the equation from there. It may be necessary to manipulate an equation further if you end up with fractions, but all you will need to do is multiply by an integer if that is the case.

First, we will balance the carbons. There are three carbons in propane, so we will make sure there are three carbons on the right side of the arrow as well:

$C_3H_8 + O_2 \rightarrow 3CO_2 + H_2O$

This is not complete. We will next balance the hydrogens. There are eight hydrogens on the left side of the equation, so:

$C_3H_8 + O_2 \rightarrow 3CO_2 + 4H_2O$

The last step is to balance the oxygens on the left and right side of the equation

$C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$

Our equation is balanced and all coefficients are integers. If we begin with one mole of propane, we will produce four moles of water.

Compare your answer with the correct one above

Question

Consider the following unbalanced equation:

$Fe+Cl_2\rightarrow FeCl_3$

How many grams of solid iron are needed to make 36.0g of $FeCl_3$ ? Assume that chlorine is in excess.

Answer

$Fe+Cl_2\rightarrow FeCl_3$

First, we will balance the equation:

$2Fe+3Cl_2\rightarrow 2FeCl_3$

Since chlorine is in excess, we know that the limiting reagent is iron.

$36.0\hspace{1 mm}g\hspace{1 mm}FeCl_3\times\frac{1\hspace{1 mm}mole\hspace{1 mm}FeCl_3}{162.2\hspace{1 mm}g\hspace{1 mm}FeCl_3}\times\frac{2\hspace{1 mm}moles\hspace{1 mm}Fe}{2\hspace{1 mm}moles\hspace{1 mm}FeCl_3}\times\frac{55.845\hspace{1 mm}g\hspace{1 mm}Fe}{1\hspace{1 mm}mole\hspace{1 mm}Fe}=12.4\hspace{1 mm}g\hspace{1 mm}Fe$

Compare your answer with the correct one above

Question

Consider the following reaction:

$KMnO_{4} + HCl \rightarrow KCl + MnCl_{2} + H_{2}O + Cl_{2}$

When the equation is balanced, what will be the coefficient in front of HCl?

Answer

When balancing equations, the goal is to make sure that the same atoms, in both type and amount, are on both the reactant and product side of the equation. A helpful approach is to write down the number of atoms already on both sides of the unbalanced equation. This way, you can predict which compounds need to be increased on which side in order to balance the equation. It also helps to balance oxygen and hydrogen last in the equation.

In this reaction, we can balance as follows.

Reactants: 1K, 1Mn, 1Cl, 4O, 1H

Products: 1K, 1Mn, 5Cl, 1O, 2H

So, we will need to increase H2O and HCl. The final balanced equation is written below.

$2KMnO_{4} + 16HCl \rightarrow 2KCl + 2MnCl_{2} + 8H_{2}O + 5Cl_{2}$

Compare your answer with the correct one above

Question

Balance the following chemical equation.

$NaOH + FeCl_{3} \rightarrow NaCl + Fe(OH)_3$

Answer

To balance an equation, we need to make sure there is the same amount of elements to the left of the arrow as there is to the right. We also need all the charges to balance out. We notice right away that there are three chlorine atoms on the left, but only one on the right.

$NaOH + FeCl_{3} \rightarrow NaCl + Fe(OH)_3$ (1Na, 1O, 1H, 1Fe, 3Cl : 1Na, 3O, 3H, 1Fe, 1Cl)

We can solve this by multiplying NaCl by three.

$NaOH + FeCl_{3} \rightarrow 3NaCl + Fe(OH)_3$ (1Na, 1O, 1H, 1Fe, 3Cl : 3Na, 3O, 3H, 1Fe, 3Cl)

This causes us to have an imbalance of sodium, which we can correct by manipulating NaOH.

$3NaOH + FeCl_{3} \rightarrow 3NaCl + Fe(OH)_3$ (3Na, 3O, 3H, 1Fe, 3Cl : 3Na, 3O, 3H, 1Fe, 1Cl)

This is the final balanced equation. Note that it is usually easiest to manipulate oxygen and hydrogen last, since they are often involved in multiple molecules.

Compare your answer with the correct one above

Question

Calcium hydroxide is treated with hydrochloric acid to produce water and calcium chloride. Write a balanced chemical reaction that describes this process.

Answer

Calcium is in the second group of the periodic table, and is therefore going to have a $\small +2$ oxidation number. Hydroxide ions have a $\small -1$ charge. Calcium hydroxide will have the formula $\small Ca(OH)_2$ .

Chloride ions have a $\small -1$ charge and hydrogen ions have a $\small +1$ charge. The formula for hydrochloric acid is $\small HCl$ .

On the products side, water has the formula $\small H_2O$ and calcium chloride has the formula $\small CaCl_2$ .

Now that we know all of the formulas, we can write our reaction:

$Ca(OH)_2+HCl\rightarrow H_2O+CaCl_2$

In order to balance the chloride atoms, we need to add coefficients.

$Ca(OH)_2+2HCl\rightarrow 2H_2O+CaCl_2$

Compare your answer with the correct one above

Question

$S + HNO_{3} \rightleftharpoons H_{2}SO_{4} + NO_{2} + H_{2}O$

In the balanced version of the preceding equation, what is the coefficient of nitrogen dioxide?

Answer

In the balanced version of the equation:

$S + 6HNO_{3} \rightleftharpoons H_{2}SO_{4} + 6NO_{2} + 2H_{2}O$

Nitrogen dioxide's coefficient is 6 in order to balance the 6 moles of nitrogen provided by the 6 moles of nitric acid on the equation's left side.

Compare your answer with the correct one above

Question

Transesterification is an industrially important process for the production of biodiesel fuel (that most diesel engines can run cleanly on) and glycerin from triglycerides (usually people use vegetable oil as their feedstock).

One easy to do and high yielding method involves the reaction of ethanol or methanol with sodium hydroxide to produce the super bases sodium ethoxide or sodium methoxide respectively. The super base is then used in the transesterification reaction with triglycerides to produce glycerin and ethyl or methyl esters (biodiesel). The products can easily be separated by differences in density. Note: For this reaction to occur the reactants must be free of water as when water is present saponification occurs (how soap is made) which will compete with the transesterification reaction. Note: just because the reaction described above is easy doesn't mean that it is anywhere near safe; please do thoroughly research any reaction you plan to undertake and take all possible safety precautions. That being said, this process is becoming much more popular lately and with the recent focus on alternative energy may well soon account for a decent portion of fuel production.

Catalyst formation equation:

$NaOH+CH_3OH\rightleftharpoons Na^+CH_3OH+H_2O$

Unbalanced transesterification equation:

$RCO_2CH_2CH(O_2CR)CH_2CO_2R +CH_3OH \xrightarrow[]{Catalyst} C_3H_8O_3+CH_3CO_2R$

Balance the following transesterification reaction assuming that all R-groups are identical:

$RCO_2CH_2CH(O_2CR)CH_2CO_2R +CH_3OH \xrightarrow[]{Catalyst} C_3H_8O_3+CH_3CO_2R$

Answer

The steps for balancing equations are as follows and should be done in order:

1.) Check for diatomic molecules: In this case there are none so move on to step 2

2.) Balance the metals (hydrogen is not considered a metal in this application): We have no metals in our equation, so move on to step 3 (the sodium is part of the catalyst so we don't consider it part of the balancing).

3.) Balance the nonmetals (not including oxygen). If you notice the left hand side of the equation has 3R groups and the right hand side only has one R-group. To fix this we can put a coefficient of 3 in front of on the right hand side giving us the following:

$RCO_2CH_2CH(O_2CR)CH_2CO_2R +CH_3OH \xrightarrow[]{Catalyst} C_3H_8O_3+3CH_3CO_2R$

On the left hand side of the equation we have 7 carbon atoms but on the right hand side we have 9 carbon atoms. This can be rectified by putting a coefficient of 3 in front of on the left hand side of the equation giving us the following:

$RCO_2CH_2CH(O_2CR)CH_2CO_2R +3CH_3OH \xrightarrow[]{Catalyst} C_3H_8O_3+3CH_3CO_2R$

Now all the nonmetals (excluding oxygen) have been balanced and we can move on to step 4

4.) Balance oxygen. There are 9 oxygen atoms on both sides of the equation. Move on to step 5

5.) Balance Hydrogen. There are 17 hydrogens on both sides of the equation.

6.) Recount all atoms to make sure you have balanced correctly. 9 carbons on both sides, 3 R-groups on both sides, 9 oxygen on both sides, and 17 hydrogens on both sides.

If this problem had involved ionic species you would also want to make certain that the charges are balanced on both sides of the equation.

The charges on both sides add up to zero so your equation is balanced:

$RCO_2CH_2CH(O_2CR)CH_2CO_2R +3CH_3OH \xrightarrow[]{Catalyst} C_3H_8O_3+3CH_3CO_2R$

Compare your answer with the correct one above

Question

Consider the following unbalanced reaction:

$AgNO_{3}+K_{3}PO_{4}\rightarrow Ag_{3}PO_{4}+KNO_{3}$

Once this equation has been balanced, what are the the respective stoichiometric coefficients (listed in the order in which the above compounds appear in the reaction)?

Answer

To balance chemical reactions, it's usually more effective to save the more common elements, such as oxygen, for last. For this reaction, let's go ahead and start by balancing the element on both sides.

$3AgNO_{3}+K_{3}PO_{4}\rightarrow Ag_{3}PO_{4}+KNO_{3}$

Because there are moles of on the right side, we'll need to have moles of $AgNO_{3}$ . Next, let's go ahead and balance nitrogen on both sides.

$3AgNO_{3}+K_{3}PO_{4}\rightarrow Ag_{3}PO_{4}+3KNO_{3}$

Since we added a coefficient of to the $AgNO_{3}$ in the first step, there are now moles of on the left. Thus, we'll need to add a coefficient of to the $KNO_{3}$ . Next, if we check the amount of , , and on both sides, we see that they are equal. Thus, we can recognize that $K_{3}PO_{4}$ and $Ag_{3}PO_{4}$ have coefficients of , and do not need to be altered.

Our final balanced equation looks like this.

$3AgNO_{3}+1K_{3}PO_{4}\rightarrow 1Ag_{3}PO_{4}+3KNO_{3}$

Compare your answer with the correct one above

Question

Consider the following balanced reaction:

2Ca(s) + O2(g) → 2CaO

Consider the reaction above. If 60g of Ca is placed in a reaction chamber with 32g of O2 which will be the limiting reactant?

Answer

When considering Limiting Reactant problems the most important aspect to consider is the molar ratio of the reactants. Here the balanced formula tells us that for every 2 moles of Ca there must be 1 mole of O2 to create the product. The amounts given by the problem are the actual amounts we are given and can be compared to the molar ratio to determine the limiting reactant. Since there is 1 mole of O2, in any situation in which there are less than 2 moles of Ca, Ca is the limting reactant. Conversely if there were 2 moles of Ca, any situation in which there less than 1 mole of O2 would be a situation in which O2 is the limiting reactant.

Compare your answer with the correct one above

Question

Given the reaction

H2SO4 (aq) + 2 LiOH (aq) → Li2SO4 (aq) + 2 H2O (l)

If you have 100g of H2SO4 and 65g of LiOH. What is your limiting reactant?

Answer

H2SO4 molecular weight is 98. This gives 1.02moles. Since only 1 mole is needed per reaction it allows the reaction to go through 1.02 times.

LiOH molecular weight is 24. This gives 2.71 moles. Every reaction requires 2 moles so the reaction can go through 1.35 times.

Based on this, H2SO4 is the limiting reactant.

Compare your answer with the correct one above

Question

What determines the amount of product formed in an irreversible reaction?

Answer

In irreversible reactions, the reaction proceeds in one direction only and these reactions usually go to completion. Since the forward reaction is the only one that occurs, the amount of product formed is only based on the reactants present, namely, the amount of limiting reactant.

Compare your answer with the correct one above

Question

Consider the following reaction:

$8Fe\hspace{1 mm}+S_8\hspace{1 mm}\rightarrow8FeS$

If we begin with 293 g Fe and 17.2 g $S_8$ , how many grams of $FeS$ will we create?

Answer

First, calculate the theoretical yield as if we had 293 g $Fe$ and excess $S_8$ :

$293\hspace{1 mm}g\hspace{1 mm}Fe\times\frac{1\hspace{1 mm}mole\hspace{1 mm}Fe}{55.8\hspace{1 mm}g\hspace{1 mm}Fe}\times\frac{8\hspace{1 mm}moles\hspace{1 mm}FeS}{8\hspace{1 mm}moles\hspace{1 mm}Fe}=5.25\hspace{1 mm}mol\hspace{1 mm}FeS$

Then calculate the theoretical yield as if we had 17.2 g $S_8$ :

$17.2\hspace{1 mm}g\hspace{1 mm}S_8\times\frac{1\hspace{1 mm}mole\hspace{1 mm}S_8}{256.48\hspace{1 mm}g\hspace{1 mm}S_8}\times\frac{8\hspace{1 mm}moles\hspace{1 mm}FeS}{1\hspace{1 mm}mole\hspace{1 mm}S_8}=5.36\times10^{-1}\hspace{1 mm}moles\hspace{1 mm} FeS$

The limiting reagent is $S_8$ as the theoretical yield calculation is lower. Now all there is to do is convert moles $FeS$ into grams:

$5.36\times10^{-1}\hspace{1 mm}moles\hspace{1 mm}FeS\times\frac{87.91\hspace{1 mm}g\hspace{1 mm}FeS}{1\hspace{1 mm}mole\hspace{1 mm}FeS}=47.1\hspace{1 mm}g\hspace{1 mm}FeS$

Compare your answer with the correct one above

Question

Consider the following reaction:

$2Na+2H_2O\rightarrow 2NaOH+H_2$

If you have 36 grams of $Na$ and 53 mL $H_2O$ , will you have any remaining sodium and/or water? How much of each?

Answer

First, let us consider the 36 grams of solid sodium

$36\hspace{1 mm}g\hspace{1 mm}Na\times\frac{1\hspace{1 mm}mole\hspace{1 mm}Na}{22.99\hspace{1 mm}g\hspace{1 mm}Na}\times\frac{2\hspace{1 mm}moles\hspace{1 mm}H_2O}{2\hspace{1 mm}moles\hspace{1 mm}Na}\times\frac{18\hspace{1 mm}g\hspace{1 mm}H_2O}{1\hspace{1 mm}mole\hspace{1 mm}H_2O}\times\frac{1\hspace{1 mm}mL\hspace{1 mm}H_2O}{1\hspace{1 mm}g\hspace{1 mm}H_2O}=28.2\hspace{1 mm}mL\hspace{1 mm}H_2O$

We now know that sodium is the limiting reagent as it uses only 28.2 mL of water and we have 53 mL of water. We will have no remaining sodium.

We will now calculate the remaining volume of water

$53\hspace{1 mm}mL-28.2\hspace{1 mm}mL=24.8\hspace{1 mm}mL\hspace{1 mm}H_2O$

Compare your answer with the correct one above

Question

The formation of ammonia is given by the following equation:

$3H_2+N_2\rightarrow 2NH_3$

If you start the reaction with 12g of hydrogen gas and 64g of nitrogen gas, which will be the limiting reagent?

Answer

When finding the limiting reactant, it is a good idea to take one of the reactants and solve for how much of the other will be needed to fully react it. For example, let us calculate how much nitrogen gas we would need in order to use up all 12g of hydrogen gas. Using the equation to find moles, we find that we have 6mol of hydrogen gas.

$12gH_2\frac{1molH_2}{2gH_2}=6molH_2$

Using the molar ratio of 1:3 from the chemical equation, we conclude that we only need 2mol of nitrogen gas to use up all of the hydrogen gas.

$6molH_2\frac{1molN_2}{3molH_2}=2molN_2$

Multiplying 2mol of nitrogen gas by its molar mass (28 grams per mol) gives us 56g of nitrogen gas needed to react all of the hydrogen.

$2molN_228\frac{g}{mol}N_2=56gN_2$

Since we started with 64g of nitrogen gas, we know that we have more than enough and that hydrogen will be used up first.* Hydrogen gas is the limiting reactant.

Compare your answer with the correct one above

Question

Magnesium will combine with oxygen gas to form magnesium oxide according to the balanced equation below.

$2Mg + O_{2}\rightarrow2MgO$

60 grams of magnesium metal and 30 grams of oxygen gas are allowed to react. If the reaction runs to completion, which reactant will be depleted first?

Answer

The reactant that will be depleted first is called the limiting reagent. We can determine the limiting reagent by calculating how much oxygen gas is necessary to use up all 60 grams of magnesium.

We start by converting the magnesium to moles.

$60g\ Mg\frac{1mol\ Mg}{24.3g\ Mg}=2.47mol\ Mg$

We then compare the molar ratio of magnesium to oxygen gas. Since we need 1 mol of oxygen gas for every 2 moles of magnesium, we only need 1.24 moles of oxygen gas to fully react the magnesium.

$2.47mol\ Mg\frac{1mol\ O_2}{2mol\ Mg}=1.23mol\ O_2$

Multiplying by the molar mass, we determine that we need 39.51 grams of oxygen gas. Remeber that the oxygen is diatomic in its gaseous state.

$1.23mol\ O_2*\frac{32g\ O_2}{1mol\ O_2}=39.51g\ O_2$

Since we only have 30 grams in the givevn question, we conclude that oxygen gas will be depleted first if the reaction runs to completion.

Compare your answer with the correct one above

Question

Consider the following balanced reaction.

$Cl_{2(g)} + 2KI_{(aq)} \rightarrow 2KCl_{(aq)} + I_{2}_{(s)}$

If the reaction starts with 8 moles of chlorine gas and 10 moles of potassium iodide, how many moles of the excess reagent will be left over after the reaction has run to completion?

Answer

When reading the balanced reaction, you should notice that it requires 1 mol of chlorine gas and 2 moles of potassium iodide in order to react and create the products. This ratio is seen in the coefficients of the reagents.

Using this ratio, we can calculate the mole of chlorine needed to react with 10 moles of potassium iodide.

$10mol\ KI*\frac{1mol\ Cl_2}{2mol\ KI}=5mol\ Cl_2$

5 moles of chlorine gas will be used to react all of the potassium iodide. Since we started with 8 moles of chlorine gas, this means that there are 3 moles of chlorine gas left over after the reaction has run to completion.

Compare your answer with the correct one above

Tap the card to reveal the answer