91Ó°ÊÓ

Consider the equation \ \( \mathrm{Zn}(s)+\mathrm{HNO}_{3}(a q) \rightarrow \mathrm{H}_{2}(g)+\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}(a q) \) \ Start by balancing the elements that appear in only one reactant and one product. Here, we balance Zn first. \ The reactant side has one Zn and the product side has one Zn, so it's already balanced. \ Next, balance the hydrogen by making sure there are the same number of hydrogens on both sides: \( 2 \) H atoms in \( \mathrm{H}_{2}(g) \) and \( 2 \) H atoms in \( 2 \) \(\mathrm{HNO}_{3}(a q)\). \ Adjust the number of \( \mathrm{HNO}_{3} \) to balance the H atoms: \(\mathrm{2HNO}_{3}(a q)\). \ Finally, adjust the nitrates (\( \mathrm{NO}_{3} \)): \ \( \mathrm{Zn}(s) + 2\mathrm{HNO}_{3}(a q) \rightarrow \mathrm{H}_{2}(g) + \mathrm{Zn}(\mathrm{NO}_{3})_{2}(a q) \) is now balanced.

Consider the equation \ \(\mathrm{Al}(s) + \mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{H}_{2}(g) + \mathrm{Al}_{2}(\mathrm{SO}_{4})_{3}(a q)\) \ First balance Al. The product side has 2 Al, so the reactant side needs to have 2 Al: \(2\mathrm{Al}(s)\). \ Next balance the sulfate ions (\( \mathrm{SO}_{4} \)): \( 3 \) sulfates on the product side so there should be \(3 \mathrm{H}_{2}\mathrm{SO}_{4}(a q)\). \ Finally, balance the hydrogen atoms: \( 6 \) H atoms in \( 3 \mathrm{H}_{2}\mathrm{SO}_{4}(a q) \) gives \( \mathrm{3H}_{2}(g) \). \ The balanced equation is: \ \( 2 \mathrm{Al}(s) + 3 \mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow 3 \mathrm{H}_{2}(g) + \mathrm{Al}_{2}(\mathrm{SO}_{4})_{3}(a q) \)

Consider the equation \ \( \mathrm{K}_{2} \mathrm{SO}_{4}(a q) + \mathrm{BaCl}_{2}(a q) \rightarrow \mathrm{BaSO}_{4}(s) + \mathrm{KCl}(a q) \) \ Start by balancing the Ba and SO4 ions: \( 1 \) Ba on each side, and \( 1 \) SO4 on each side. \ Next, balance the K and Cl atoms. There are \( 2 \) K atoms in \(\mathrm{K}_{2} \mathrm{SO}_{4} \), so we need \(2\mathrm{KCl}(a q)\) on the product side. \ The final balanced equation is: \ \( \mathrm{K}_{2}\mathrm{SO}_{4}(a q) + \mathrm{BaCl}_{2}(a q) \rightarrow \mathrm{BaSO}_{4}(s) + 2\mathrm{KCl}(a q) \)

Consider the equation \ \( \mathrm{CaCO}_{3}(s) \rightarrow \mathrm{CaO}(s) + \mathrm{CO}_{2}(g) \) \ Since each element appears in only one reactant and one product, check them one by one. \ There is \( 1 \) Ca atom on both sides, \( 1 \) C atom on both sides, and \( 3 \) O atoms on both sides. \ The equation is already balanced: \( \mathrm{CaCO}_{3}(s) \rightarrow \mathrm{CaO}(s) + \mathrm{CO}_{2}(g) \)