MCAT Physical - Reaction Calculations and Limiting Reagent

$2Mn + 10HCl \rightarrow 2MnCl_{5} + 5H_{2}$

5.6 grams of manganese reacts with 650 mL of 6.0 M hydrochloric acid to form manganese (V) chloride and hydrogen gas. Along with the products, a large amount of heat is evolved.

Assuming standard temperature and pressure, what volume of hydrogen gas is produced by this reaction?

Explanation

Convert 5.6g of manganese (limiting reagent) to volume of hydrogen gas.

$5.6 g Mn * (\frac{1 mol Mn}{54.9 g Mn}) * (\frac{5 mol H_{2}}{2 mol Mn}) * (\frac{22.4 L H_{2}}{1 mol H_{2}}) = 5.7 L$

Which of the following produces a $\small 0.1 M$ solution of potassium chloride?

Put $\small 0.746g$ of solute into a container and bring the volume to $\small 100mL$ by adding water while stirring

Add $\small 74.6g$ of solute to one liter of water

Add $\small 7.46g$ of solute to $\small 100mL$ of water

Put $\small 147.51g$ of solute into a container and bring the volume to one liter by adding water while stirring

Potassium chloride is not soluble in water

Explanation

We will be looking for a solution that results in one mole of potassium chloride per ten liters of water.

$0.1M=\frac{0.10mol}{1L}$

We will need to find the molecular weight of potassium chloride.

$KCl\rightarrow MW=39.1+35.5=74.6\frac{g}{mol}$

In order to get the desired concentration, we will need to add one-tenth of this amount to one liter of water.

$7.46g*\frac{1mol}{74.6g}=0.1mol$

Our ratio, then is:

$\frac{7.46g}{1L}$

The only answer to follow this ratio is $\small 0.746g$ of potassium chloride in $\small 100mL$ .

$\frac{7.46g\div10}{1L\div 10}=\frac{0.746g}{0.1L}$

What products will be formed by the reaction between copper (II) fluoride and sodium sulfate?

CuSO4 and NaF

Cu2SO4 and NaF

CuS and NaF

Cu2Na and F2SO4

Explanation

Recognize that this is a double replacement reaction in which ion pairs will switch. After reaction, copper (II) will now be paired with sulfate, and sodium with flouride according to the (unbalanced) equation below. Remember that sulfate has a charge of -2, and flouride of -1.

$CuF_{2 } + Na_{2}SO_{4}\rightarrow CuSO_{4} + NaF$

Which represents the correct balanced equation for the reaction between silver (I) nitrate and magnesium hydroxide?

$2Ag(NO_{3}) + Mg(OH)_{2} \rightarrow Mg(NO_{3})_{2} + 2AgOH$

$2Ag_{3}N + 3Mg(OH)_{2} \rightarrow Mg_{}3N_{}2 + 6AgOH$

$2Ag_{2}(NO_{3}) + Mg(OH)_{2} \rightarrow 2Ag_{2}(OH) + Mg(NO_{3})_{2}$

$2Ag(NO_{3}) + Mg(OH)_{2} \rightarrow Mg(NO_{3})_{2} + AgOH$

$2Ag(NO_2)+Mg(OH)_2\rightarrow Mg(NO_2)_2+2AgOH$

Explanation

Silver (I) nitrate is AgNO3. Recognizing this allows us to eliminate two answer choices, which incorrectly substitute nitrogen (N) for nitrate (NO3) or balance the molecular charges incorrectly. Of the two remaining choices, only one is balanced correctly.

Hydrogen can be produced on a large scale by a method called the Bosch process, by which steam is passed over heated iron shavings. The reaction is shown below:

$3Fe(s) + 4H_{2}O(g) \rightarrow Fe_{3}O_{2}(s) + 4 H_{2}(g)$

Alternatively, hydrogen can be produced by reacting steam with natural gas, according to the following equation:

$CH_{4}(g) + H_{2}O(g) \rightarrow CO(g) + 3 H_{2}(g)$

How many molecules of carbon monoxide are produced per liter of hydrogen, when using the natural gas method of production at STP?

$9*10^{21}\ \text{molecules}$

$5*10^{24}\ \text{molecules}$

$6*10^{21}\ \text{molecules}$

$3*10^{22}\ \text{molecules}$

Explanation

Since the reaction takes place at STP, we can convert liters of hydrogen to moles using a constant.

$1L\ H_2*\frac{1mol}{22.4L}=0.0446mol\ H_2$

Using this value, we can find the moles of carbon monoxide produced and convert to molecules by using Avogadro's number.

$0.0446mol\ H_2*\frac{1mol\ CO}{3mol\ H_2}*\frac{6.022*10^{23}molec}{1mol}=9*10^{21}\ \text{molecules}\ CO$

$AlCl_{3}+3AgNO_{3}\rightarrow: Al(NO_{3})_{3}+3AgCl$

If you begin with $\small 66.67g$ of aluminum chloride and unlimited silver nitrate, how many grams of silver chloride can be produced?

$215g\ AgCl$

$35.8g\ AgCl$

$186g\ AgCl$

$201g\ AgCl$

$116g\ AgCl$

Explanation

This is a stoichiometry question requiring us to convert between grams, moles, reactants, and products.

Use the periodic table to find the molar masses of the two compounds in question.

$MW_{AlCl_3}=133.5\frac{g}{mol}$

$MW_{AgCl}=143.37\frac{g}{mol}$

We can use the reaction formula to find the ratio of aluminum chloride to silver chloride. In this case, the ratio is 1:3.

$AlCl_{3}+3AgNO_{3}\rightarrow: Al(NO_{3})_{3}+3AgCl$

Now we can set up a calculation to convert grams of aluminum chloride to grams of silver chloride, making sure that all units cancel appropriately.

$66.67g\ AlCl_3*\frac{1mol\ AlCl_3}{133.5g\ AlCl_3}*\frac{3mol\ AgCl}{1mol\ AlCl_3}*\frac{143.37g\ AgCl}{1mol\ AgCl}=214.80g\ AgCl$

Five kilograms of oxygen are consumed in a chemical reaction that generates two photons per oxygen molecule. How many photons were generated?

$1.88*10^{26\ }\text{photons}$

$5.32*10^{26\ }\text{photons}$

$3.76*10^{26\ }\text{photons}$

$9.41*10^{26\ }\text{photons}$

$4.40*10^{26\ }\text{photons}$

Explanation

First, use Avogadro's law and the molar weight of oxygen to determine the number of oxygen atoms.

$g*\frac{mol}{g}*(6.02*10^{23}\frac{moelcule}{mol})=molecule$

$5000g\ O_2*\frac{1mol\ O_2}{32g\ O_2}*(6.02*10^{23}\frac{molecules}{mol})=9.41*10^{25}\text{molecules}$

We know that two photons are formed from every oxygen molecule. We can use basic stoichiometry to find the number of photons generated.

$2\frac{photons}{molecule}(9.41*10^{25}molecule)=1.88*10^{26}\text{photons}$

Hydrogen can be produced on a large scale by a method called the Bosch process, by which steam is passed over heated iron shavings. The reaction is shown below:

$3Fe(s) + 4H_{2}O(g) \rightarrow Fe_{3}O_{2}(s) + 4 H_{2}(g)$

Alternatively, hydrogen can be produced by reacting steam with natural gas, according to the following equation:

$CH_{4}(g) + H_{2}O(g) \rightarrow CO(g) + 3 H_{2}(g)$

Suppose equal masses of iron and methane were available, along with excess water. What is the difference in the amount of hydrogen molecules that can be produced?

Methane will produce about eight times more hydrogen molecules

Iron will produce about eight times more hydrogen molecules

Iron will produce about 1.3 times more hydrogen molecules

Both methods will produce equal amounts of hydrogen molecules

Explanation

To determine which method will produce more hydrogen molecules, convert equal masses of each reactant into molecules of hydrogen gas. The conversion requires us to convert the initial compound to moles, multiply by the molar ratio from the reaction to find moles of hydrogen, and multiply by Avogadro's number to convert to molecules.

$3Fe(s) + 4H_{2}O(g) \rightarrow Fe_{3}O_{2}(s) + 4 H_{2}(g)$

$1 g Fe * \frac{1 mol Fe}{55.8 g Fe} * \frac{4 mol H_{2}}{3 mol Fe} * \frac{6.022 * 10^{23} molec H_{2}}{mol H_{2}} = 0.144 * 10^{23} molec$

$CH_{4}(g) + H_{2}O(g) \rightarrow CO(g) + 3 H_{2}(g)$

$1 g CH_{4} * \frac{1 mol CH_{4}}{16.0 g CH_{4}} * \frac{3 mol H_{2}}{1 mol CH_{4}} * \frac{6.022 * 10^{23} molec H_{2}}{mol H_{2}} = 1.14 * 10^{23} molec$

Dividing these solutions, we see that the methane method produces roughly eight times the amount of hydrogen molecules.

$\frac{1.14*10^{23}}{0.144*10^{23}}=7.9\approx 8$

When sodium nitrate reacts with iron (III) bromide to produce iron (III) nitrate, what is the molar ratio of iron (III) nitrate to sodium bromide?

1:3

1:1

1:2

3:1

Explanation

The first step is writing, and then balancing, the chemical equation for this reaction.

$3NaNO_{3} + FeBr_{3} \rightarrow Fe(NO_{3})_{3} + 3NaBr$

From this, we are able to see that for every one mole of iron (III) nitrate that is produced, three moles of sodium bromide are produced (a 1 to 3 ratio).

A researcher runs an acid-base reaction using 3g of hydrochloric acid and 5g of cesium hydroxide. He wants to use the excess cesium hydroxide for another reaction. How much excess cesium hydroxide can he obtain after the completion of the reaction?

Explanation

The first step is to determine the limiting reagent. To do this we need to first calculate the moles of reactants. The MW of hydrochloric acid is $36.46\frac{g}{mol}$ and MW of cesium hydroxide is $149.9\frac{g}{mol}$ . The moles of each reactant is

$3g:HCl \left(\frac{1: mol}{36.46g}\right) = 0.082:mol:HCl$

$5g:CsOH \left(\frac{1: mol}{149.9g}\right) = 0.033:mol:CsOH$

Next step is to write out the balanced chemical reaction

$HCl + CsOH \rightleftharpoons CsCl + H_2O$

The ratio of reactants is 1:1; therefore, since has the smaller amount of moles it is the limiting reagent. This means that all of cesium hydroxide will be utilized in this reaction and the researcher won’t be able to salvage any cesium hydroxide. Note that there will be excess left after completion of reaction. of will react with of ; therefore, there will be a total of of excess .

Reaction Calculations and Limiting Reagent

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