Chapter 3 - Equations, the Mole, and Chemical Formulas - Questions and Exercises - Exercises - Page 131: 3.24

(a) 4Al$(s)$ + 3O$_2$$(g)$ ➔ 2Al$_2$O$_3$$(s)$ (b) N$_2$$(g)$ + 3H$_2$$(g)$ ➔ 2NH$_3$$(g)$ (c) 2C$_6$H$_6$$(l)$ + 15O$_2$$(g)$ ➔ 6H$_2$O$(l)$ + 12CO$_2$$(g)$

(a) Al$(s)$ + O$_2$$(g)$ ➔ Al$_2$O$_3$$(s)$ We first look at the oxygen atoms. We have 2 on the left and 3 on the right. If we add a 3 in front of the O$_2$ and a 2 in front of Al$_2$O$_3$ to balance the oxygen. Al$(s)$ + 3O$_2$$(g)$ ➔ 2Al$_2$O$_3$$(s)$ We now look at the aluminum atoms. We have 1 on the left and 4 on the right. We place a 4 in front of the Al on the left side to balance the aluminum. 4Al$(s)$ + 3O$_2$$(g)$ ➔ 2Al$_2$O$_3$$(s)$ Now let's check the numbers of atoms in the reactants and the products. Reactants/Products: aluminum: 4/4 oxygen: 6/6 (b) N$_2$$(g)$ + H$_2$$(g)$ ➔ NH$_3$$(g)$ We look first at the hydrogen atoms. We have 2 on the left and 3 on the right. We add 3 as a coefficient in front of the H$_2$ on the left and a 2 in front of the NH$_3$ on the right to balance the hydrogen atoms. N$_2$$(g)$ + 3H$_2$$(g)$ ➔ 2NH$_3$$(g)$ We now look at the nitrogen atoms. We have 2 on the left and 2 on the right. We are balanced in terms of nitrogen. Now let's check the numbers of atoms in the reactants and the products. Reactants/Products: nitrogen: 2/2 hydrogen: 6/6 (c) C$_6$H$_6$$(l)$ + O$_2$$(g)$ ➔ H$_2$O$(l)$ + CO$_2$$(g)$ We first look at the hydrogen atoms. We have 6 on the left and 2 on the right. We need to add a 3 in front of H$_2$O to balance the hydrogen. C$_6$H$_6$$(l)$ + O$_2$$(g)$ ➔ 3H$_2$O$(l)$ + CO$_2$$(g)$ Now we look at the carbon atoms. We have 6 on the left and 1 on the right. If we add a 6 in front of CO$_2$, we will be balanced with respect to carbon. C$_6$H$_6$$(l)$ + O$_2$$(g)$ ➔ 3H$_2$O$(l)$ + 6CO$_2$$(g)$ Next, we look at the oxygen atoms. We have 2 on the left and 15 on the right. We need to add a 15 in front of the O$_2$ and change the coefficient on the H$_2$O on the right from 3 to 6 and the coefficient on the CO$_2$ from 6 to 12 to balance the oxygen. C$_6$H$_6$$(l)$ + 15O$_2$$(g)$ ➔ 6H$_2$O$(l)$ + 12CO$_2$$(g)$ Now we are imbalanced in terms of carbon. We have 6 on the left and 12 on the right. We need to add a 2 in front of the C$_6$H$_6$. This also takes care of the imbalance of hydrogen that was created (6 on the left and 12 on the right). 2C$_6$H$_6$$(l)$ + 15O$_2$$(g)$ ➔ 6H$_2$O$(l)$ + 12CO$_2$$(g)$ Now let's check the numbers of atoms in the reactants and the products. Reactants/Products: carbon: 12/12 hydrogen: 12/12 oxygen: 30/30