Question

practice: balancing equations #2
show all work on a separate sheet of paper. place coefficients of 1 blank. for 14 and 15, you must write out the equa
1 n₂ + h₂ → __ nh₃
2 kclo₃ → kcl + __ o₂
3 nacl + f₂ → naf + cl₂
4 h₂ + o₂ → __ h₂o
5 pb(oh)₂ + hcl → h₂o + pbcl₂
6 albr₃ + k₂so₄ → kbr + al₂(so₄)₃
7 ch₄ + o₂ → co₂ + h₂o
8 c₃h₈ + o₂ → co₂ + h₂o
9 fecl₃ + naoh → fe(oh)₃ + nacl
10 p + o₂ → __ p₂o₅
11 co₂ + h₂o → c₆h₁₂o₆ + o₂
12 s₈ + o₂ → __ so₃
13 mn + hi → h₂ + mni₃

Explanation:

To solve the problem of balancing the chemical equation \( \text{Pb(OH)}_2 + \text{HCl}
ightarrow \text{H}_2\text{O} + \text{PbCl}_2 \), we follow these steps:

Step 1: List the number of each atom on both sides

• Reactants (left side):
• \( \text{Pb} \): 1 (from \( \text{Pb(OH)}_2 \))
• \( \text{O} \): 2 (from \( \text{Pb(OH)}_2 \))
• \( \text{H} \): 2 (from \( \text{Pb(OH)}_2 \)) + 1 (from \( \text{HCl} \)) = 3 (but we'll track per formula unit)
• \( \text{Cl} \): 1 (from \( \text{HCl} \))
• Products (right side):
• \( \text{Pb} \): 1 (from \( \text{PbCl}_2 \))
• \( \text{O} \): 1 (from \( \text{H}_2\text{O} \))
• \( \text{H} \): 2 (from \( \text{H}_2\text{O} \))
• \( \text{Cl} \): 2 (from \( \text{PbCl}_2 \))

Step 2: Balance the chlorine atoms

To balance \( \text{Cl} \), we need 2 \( \text{HCl} \) (since \( \text{PbCl}_2 \) has 2 Cl atoms). So we put a coefficient of 2 in front of \( \text{HCl} \):
\( \text{Pb(OH)}_2 + 2\text{HCl}
ightarrow \text{H}_2\text{O} + \text{PbCl}_2 \)

Now, update the atom counts:

• Reactants:
• \( \text{Pb} \): 1
• \( \text{O} \): 2
• \( \text{H} \): 2 (from \( \text{Pb(OH)}_2 \)) + 2 (from \( 2\text{HCl} \)) = 4
• \( \text{Cl} \): 2 (from \( 2\text{HCl} \))
• Products:
• \( \text{Pb} \): 1
• \( \text{O} \): 1 (from \( \text{H}_2\text{O} \))
• \( \text{H} \): 2 (from \( \text{H}_2\text{O} \))
• \( \text{Cl} \): 2

Step 3: Balance the oxygen and hydrogen atoms

Now, look at \( \text{O} \) and \( \text{H} \). On the reactants, we have 2 O (from \( \text{Pb(OH)}_2 \)) and 4 H (from \( \text{Pb(OH)}_2 \) and \( 2\text{HCl} \)). On the products, we have 1 O (from \( \text{H}_2\text{O} \)) and 2 H (from \( \text{H}_2\text{O} \)). To balance O and H, we need 2 \( \text{H}_2\text{O} \) (since 2 \( \text{H}_2\text{O} \) gives 2 O and 4 H). So we put a coefficient of 2 in front of \( \text{H}_2\text{O} \):
\( \text{Pb(OH)}_2 + 2\text{HCl}
ightarrow 2\text{H}_2\text{O} + \text{PbCl}_2 \)

Now, check the atom counts:

• Reactants:
• \( \text{Pb} \): 1
• \( \text{O} \): 2
• \( \text{H} \): 2 + 2 = 4
• \( \text{Cl} \): 2
• Products:
• \( \text{Pb} \): 1
• \( \text{O} \): 2 (from \( 2\text{H}_2\text{O} \))
• \( \text{H} \): 4 (from \( 2\text{H}_2\text{O} \))
• \( \text{Cl} \): 2

All atoms are now balanced.

Answer:

The balanced equation is \( \boldsymbol{1}\text{Pb(OH)}_2 + \boldsymbol{2}\text{HCl}
ightarrow \boldsymbol{2}\text{H}_2\text{O} + \boldsymbol{1}\text{PbCl}_2 \) (coefficients: 1, 2, 2, 1)