a) \( \mathrm{HNO}_{3}+\mathrm{SnO} \rightarrow \mathrm{NO}+\mathrm{SnO}_{2}+\mathrm{H}_{2} \mathrm{O} \) b) \( \mathrm{MnO}_{2}+\mathrm{HCl} \rightarrow \mathrm{MnCl}_{2}+\mathrm{Cl}_{2}++\mathrm{H}_{2} \mathrm{O} \) c) \( \mathrm{H}_{2} \mathrm{~S}+\mathrm{HNO} \rightarrow \mathrm{S}+\mathrm{NO}+\mathrm{H}_{2} \mathrm{O} \) d) \( \mathrm{CuS}+\mathrm{O}_{2} \rightarrow \mathrm{Cu}_{2} \mathrm{O}+\mathrm{SO}_{2} \) e) \( \mathrm{I}_{2} \mathrm{O}_{5}+\mathrm{CO} \rightarrow \mathrm{I}_{2}+\mathrm{CO}_{2} \) f) \( \mathrm{SnCI}_{2}+\mathrm{HgCl} \rightarrow \mathrm{SnCl}_{4}+\mathrm{Hg}_{2} \mathrm{Cl}_{2} \) g) \( \mathrm{Zn}+\mathrm{S} \rightarrow \mathrm{ZnS} \) h) \( \mathrm{AgNO}+\mathrm{Pb} \rightarrow \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}+\mathrm{Ag} \) i) \( \mathrm{CuO}+\mathrm{NH}_{3} \rightarrow \mathrm{~N}_{2}+\mathrm{Cu}+\mathrm{H}_{2} \mathrm{O} \) j) \( \mathrm{H}_{2} \mathrm{O}_{2}+\mathrm{KMnO}_{4}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{O}_{2}+\mathrm{MnSO}_{4}+\mathrm{K}_{2} \mathrm{SO}_{4}+\mathrm{H}_{2} \mathrm{O} \)
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Beyond the Answer
Did you know that nitric acid (HNO₃), which plays a key role in reaction (a), was discovered in the 13th century and historically used in alchemy? It’s a powerful oxidizing agent that contributes to the formation of various nitrogen oxides, including the nitrogen monoxide example here. That’s right, we’re talking about the same nitric acid used in explosives, fertilizers, and even in your car's catalytic converter! If you find yourself tackling these reactions, a common mistake often occurs during balancing, especially in complex reactions like (j). Remember, understanding oxidation states can save you from miscounting atoms! Before jumping into balancing, take a moment to identify which elements undergo oxidation or reduction—a little organization goes a long way in ensuring you get the right products and coefficients. Happy balancing!
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