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

First, check the number of each atom on both sides. For Ca: 1 on left, 1 on right. For O: 1 on left ($\ce{H2O}$), 2 on right ($\ce{Ca(OH)2}$). For H: 2 on left ($\ce{H2O}$), 4 on right ($\ce{Ca(OH)2}$ has 2 H, $\ce{H2}$ has 2 H, total 4). So we need to balance O and H. Multiply $\ce{H2O}$ by 2. Now left: Ca=1, O=2, H=4. Right: Ca=1, O=2 ($\ce{Ca(OH)2}$), H=2 (from $\ce{H2O}$*2: 4 H, $\ce{Ca(OH)2}$ has 2 H, so $\ce{H2}$ should be 1? Wait, no: $\ce{Ca(OH)2}$ has 2 H, $\ce{H2}$ is 1 molecule (2 H). So total H on right: 2 + 2 = 4. O on right: 2. So after multiplying $\ce{H2O}$ by 2, left: Ca=1, O=2, H=4. Right: Ca=1, O=2, H=4. Now check Ca: 1 on both. So the balanced equation is $\ce{Ca(s) + 2H2O(l) -> Ca(OH)2(aq) + H2(g)}$.

Step2: Balance Equation 31

$\ce{KClO3(s) -> KCl(s) + O2(g)}$. Left: K=1, Cl=1, O=3. Right: K=1, Cl=1, O=2. To balance O, find LCM of 3 and 2, which is 6. So multiply $\ce{KClO3}$ by 2 (O=6) and $\ce{O2}$ by 3 (O=6). Now left: K=2, Cl=2, O=6. Right: K=1, Cl=1, O=6. So multiply $\ce{KCl}$ by 2. Now balanced: $\ce{2KClO3(s) -> 2KCl(s) + 3O2(g)}$.

Step3: Balance Equation 32

$\ce{Sb2S3(s) + O2(g) -> Sb2O3(s) + SO2(g)}$. First, balance Sb: 2 on left, 2 on right (good). S: 3 on left, 1 on right. So multiply $\ce{SO2}$ by 3. Now right: S=3, O=3 (from $\ce{Sb2O3}$) + 6 (from $\ce{SO2}$3) = 9. Left: O=2 (from $\ce{O2}$). Now balance O: left O=2, right O=9. LCM of 2 and 9? Wait, $\ce{Sb2O3}$ has 3 O, $\ce{SO2}$3 has 6 O, total 9. So we need to get O on left to 9. But $\ce{O2}$ is diatomic, so multiply $\ce{O2}$ by 9/2? No, use integers. Multiply the entire equation by 2 to eliminate fractions. Wait, let's try again. After balancing S: $\ce{Sb2S3 + O2 -> Sb2O3 + 3SO2}$. Now O on right: 3 (from $\ce{Sb2O3}$) + 6 (from 3$\ce{SO2}$) = 9. So O on left: 2x (where x is coefficient of $\ce{O2}$) = 9? No, wait, maybe balance O after S and Sb. Sb is balanced. S: 3 on left, so 3 $\ce{SO2}$. Now O: left is $\ce{O2}$, right is 3 (from $\ce{Sb2O3}$) + 32 (from $\ce{SO2}$) = 3 + 6 = 9. So we need 9 O on left, but $\ce{O2}$ is 2 per molecule, so coefficient of $\ce{O2}$ is 9/2. To make it integer, multiply all coefficients by 2: $\ce{2Sb2S3 + 9O2 -> 2Sb2O3 + 6SO2}$. Now check: Sb: 4 on left, 4 on right (22). S: 6 on left (23), 6 on right (61). O: 18 on left (92), 6 (23) + 12 (6*2) = 18 on right. Balanced.

Answer:

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