Balancing Equations - AP Chemistry

Balancing equations:

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

steps:

1. balance the Na first; get 2 NaOH

2. balance H next; get 1 H2SO4, 2 H2O

3. check to make sure that O and S balance

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

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.

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.

Fe+Cl_2rightarrow FeCl_3

First, we will balance the equation:

2Fe+3Cl_2rightarrow 2FeCl_3

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

36.0hspace{1 mm}ghspace{1 mm}FeCl_3times$\frac{1hspace{1 mm}$molehspace{1 mm}FeCl_3}{162.2hspace{1 mm}ghspace{1 mm}FeCl_3}times$\frac{2hspace{1 mm}$moleshspace{1 mm}Fe}{2hspace{1 mm}moleshspace{1 mm}FeCl_3}times$\frac{55.845hspace{1 mm}$ghspace{1 mm}Fe}{1hspace{1 mm}molehspace{1 mm}Fe}=12.4hspace{1 mm}ghspace{1 mm}Fe

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.

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.

(1Na, 1O, 1H, 1Fe, 3Cl : 1Na, 3O, 3H, 1Fe, 1Cl)

We can solve this by multiplying NaCl by three.

(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.

(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.

Calcium is in the second group of the periodic table, and is therefore going to have a oxidation number. Hydroxide ions have a charge. Calcium hydroxide will have the formula .

Chloride ions have a charge and hydrogen ions have a charge. The formula for hydrochloric acid is .

On the products side, water has the formula and calcium chloride has the formula .

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

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

In the balanced version of the equation:

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.

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:

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:

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:

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.

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

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

Our final balanced equation looks like this.

Balancing equations:

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

steps:

1. balance the Na first; get 2 NaOH

2. balance H next; get 1 H2SO4, 2 H2O

3. check to make sure that O and S balance

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

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

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.

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.

Fe+Cl_2rightarrow FeCl_3

First, we will balance the equation:

2Fe+3Cl_2rightarrow 2FeCl_3

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

36.0hspace{1 mm}ghspace{1 mm}FeCl_3times$\frac{1hspace{1 mm}$molehspace{1 mm}FeCl_3}{162.2hspace{1 mm}ghspace{1 mm}FeCl_3}times$\frac{2hspace{1 mm}$moleshspace{1 mm}Fe}{2hspace{1 mm}moleshspace{1 mm}FeCl_3}times$\frac{55.845hspace{1 mm}$ghspace{1 mm}Fe}{1hspace{1 mm}molehspace{1 mm}Fe}=12.4hspace{1 mm}ghspace{1 mm}Fe

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.

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.

(1Na, 1O, 1H, 1Fe, 3Cl : 1Na, 3O, 3H, 1Fe, 1Cl)

We can solve this by multiplying NaCl by three.

(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.

(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.