Question

balancing equations: balance the equation in t

1. ca(s) + h₂o(l) ⟶ ca(oh)₂(aq) + h₂(g)

1. kclo₃(s) ⟶ kcl(s) + o₂(g)

1. sb₂s₃(s) + o₂(g) ⟶ sb₂o₃(s) + so₂(g)

Explanation:

Step1: Balance Equation 30

• Ca atoms: 1 on left, 1 on right (balanced).
• O atoms: 1 on left ($\ce{H2O}$), 2 on right ($\ce{Ca(OH)2}$). Multiply $\ce{H2O}$ by 2: $\ce{Ca(s) + 2H2O(l) -> Ca(OH)2(aq) + H2(g)}$
• H atoms: 4 on left (2$\ce{H2O}$), 2 (from $\ce{Ca(OH)2}$) + 2 (from $\ce{H2}$) = 4 on right. Now check Ca: still 1. Equation: $\ce{Ca(s) + 2H2O(l) -> Ca(OH)2(aq) + H2(g)}$

Step2: Balance Equation 31

• K atoms: 1 on left, 1 on right (balanced).
• Cl atoms: 1 on left, 1 on right (balanced).
• O atoms: 3 on left ($\ce{KClO3}$), 2 on right ($\ce{O2}$). Find LCM of 3 and 2 (6). Multiply $\ce{KClO3}$ by 2, $\ce{O2}$ by 3: $\ce{2KClO3(s) -> 2KCl(s) + 3O2(g)}$
• Check K: 2 on left, 2 on right. Cl: 2 on left, 2 on right. O: 6 on left (2×3), 6 on right (3×2). Balanced.

Step3: Balance Equation 32

• Sb atoms: 2 on left ($\ce{Sb2S3}$), 2 on right ($\ce{Sb2O3}$) (balanced).
• S atoms: 3 on left ($\ce{Sb2S3}$), 1 on right ($\ce{SO2}$). Multiply $\ce{SO2}$ by 3: $\ce{Sb2S3(s) + O2(g) -> Sb2O3(s) + 3SO2(g)}$
• O atoms: 3 on left ($\ce{Sb2O3}$) + 6 (from 3$\ce{SO2}$) = 9 on right. On left, $\ce{O2}$: let coefficient be $x$. So $2x = 9$? Wait, recalculate: $\ce{Sb2O3}$ has 3 O, 3$\ce{SO2}$ has 6 O: total 9 O on right. So $\ce{O2}$ needs 9/2? No, use whole numbers. Multiply entire equation by 2 to eliminate fraction? Wait, original: $\ce{Sb2S3 + O2 -> Sb2O3 + 3SO2}$. O on right: 3 + 3×2 = 9. So O2 coefficient: 9/2. Multiply all by 2: $\ce{2Sb2S3(s) + 9O2(g) -> 2Sb2O3(s) + 6SO2(g)}$
• Check Sb: 4 on left (2×2), 4 on right (2×2). S: 6 on left (2×3), 6 on right (6×1). O: 18 on left (9×2), 2×3 + 6×2 = 6 + 12 = 18 on right. Balanced.

Answer:

1. $\boldsymbol{\ce{Ca(s) + 2H2O(l) -> Ca(OH)2(aq) + H2(g)}}$
2. $\boldsymbol{\ce{2KClO3(s) -> 2KCl(s) + 3O2(g)}}$
3. $\boldsymbol{\ce{2Sb2S3(s) + 9O2(g) -> 2Sb2O3(s) + 6SO2(g)}}$