Question

balancing equations: balance the equation

1. ca(s) + h₂o(l) → ca(oh)₂(aq) + h₂(g)
2. kclo₃(s) → kcl(s) + o₂(g)
3. sb₂s₃(s) + o₂(g) → sb₂o₃(s) + so₂(g)

Explanation:

Step1: Balance Equation 30 ($\ce{Ca(s) + H_{2}O(l) -> Ca(OH)_{2}(aq) + H_{2}(g)}$)

• Check O: Left has 1 O, right has 2 O. So put 2 in front of $\ce{H_{2}O}$.
• Now equation: $\ce{Ca + 2H_{2}O -> Ca(OH)_{2} + H_{2}}$
• Check H: Left has $2\times2 = 4$ H, right has $2 + 2 = 4$ H. Ca is balanced (1 on each side). So balanced equation: $\ce{Ca(s) + 2H_{2}O(l) = Ca(OH)_{2}(aq) + H_{2}(g)}$

Step2: Balance Equation 31 ($\ce{KClO_{3}(s) -> KCl(s) + O_{2}(g)}$)

• Check O: Left has 3 O, right has 2 O. Find LCM of 3 and 2, which is 6. So put 2 in front of $\ce{KClO_{3}}$ and 3 in front of $\ce{O_{2}}$.
• Now equation: $\ce{2KClO_{3} -> KCl + 3O_{2}}$
• Check K and Cl: Left has 2 K and 2 Cl, right has 1 K and 1 Cl. So put 2 in front of $\ce{KCl}$. Balanced equation: $\ce{2KClO_{3}(s) = 2KCl(s) + 3O_{2}(g)}$

Step3: Balance Equation 32 ($\ce{Sb_{2}S_{3}(s) + O_{2}(g) -> Sb_{2}O_{3}(s) + SO_{2}(g)}$)

• Check Sb: Already balanced (2 on each side).
• Check S: Left has 3 S, right has 1 S. Put 3 in front of $\ce{SO_{2}}$.
• Now equation: $\ce{Sb_{2}S_{3} + O_{2} -> Sb_{2}O_{3} + 3SO_{2}}$
• Check O: Left has 2 O, right has $3 + 3\times2 = 9$ O. Multiply $\ce{O_{2}}$ by $\frac{9}{2}$, but use integers. Multiply all by 2: $\ce{2Sb_{2}S_{3} + 9O_{2} -> 2Sb_{2}O_{3} + 6SO_{2}}$
• Check all: Sb: $2\times2 = 4$ (left), $2\times2 = 4$ (right); S: $2\times3 = 6$ (left), $6$ (right); O: $9\times2 = 18$ (left), $2\times3 + 6\times2 = 6 + 12 = 18$ (right). Balanced.

Answer:

1. $\boldsymbol{\ce{Ca(s) + 2H_{2}O(l) = Ca(OH)_{2}(aq) + H_{2}(g)}}$
2. $\boldsymbol{\ce{2KClO_{3}(s) = 2KCl(s) + 3O_{2}(g)}}$
3. $\boldsymbol{\ce{2Sb_{2}S_{3}(s) + 9O_{2}(g) = 2Sb_{2}O_{3}(s) + 6SO_{2}(g)}}$