U4.3 · Representations of ReactionsPractice · 反应的表示方式

One $\ce{O2}$ is left over: it didn’t react, so it doesn’t belong in the equation. What reacted was 4 $\ce{H2S}$ and 6 $\ce{O2}$, forming 4 $\ce{SO2}$ and 4 $\ce{H2O}$: a 4 : 6 : 4 : 4 ratio, or 2 : 3 : 2 : 2 in lowest terms. A includes the leftover $\ce{O2}$, C isn’t in lowest terms, and D doesn’t conserve atoms.

6 $\ce{CH4}$ would need 12 $\ce{O2}$, but there are only 10, so $\ce{O2}$ runs out: 10 $\ce{O2}$ react with 5 $\ce{CH4}$, forming 5 $\ce{CO2}$ and 10 $\ce{H2O}$, with 1 $\ce{CH4}$ left over. Total: 1 + 5 + 10 = 16. Don’t forget the leftover reactant is still in the container.

Na forms +1 ions, so sodium hydroxide is NaOH. Balanced: 2 Na and 2 $\ce{H2O}$ give 2 NaOH and 1 $\ce{H2}$. B has unequal numbers of H atoms; C changes a formula to balance ($\ce{Na(OH)2}$ doesn’t exist), but balancing changes coefficients, never subscripts; D has the wrong product.

Balance C and H first: 1 $\ce{C4H10}$ gives 4 $\ce{CO2}$ and 5 $\ce{H2O}$, which hold 4 × 2 + 5 = 13 O atoms, so 13/2 $\ce{O2}$. Double everything: $\ce{2C4H10(g) +}$ $\ce{13O2(g) ->}$ $\ce{8CO2(g) +}$ $\ce{10H2O(l)}$. The coefficient of $\ce{O2}$ is 13.

9 $\ce{B2}$ can react with only 9 ÷ 3 = 3 $\ce{A2}$, forming 6 $\ce{AB3}$ and leaving 1 $\ce{A2}$. Check atoms: before, 8 A and 18 B; after, A = 6 + 2 = 8 and B = 18, both conserved. None of the other choices conserves atoms.