Questions: Balance each of the following chemical equations. (Use the lowest possible coefficients for all reactions.) a. NaO2(s) + H2O(l) → NaOH(aq) + O2(g) + H2O2(aq) b. Fe2O3(s) + HBr(aq) → FeBr3(aq) + H2O(l) c. H2S(g) + O2(g) → SO2(g) + H2O(g) d. SO2Cl2(l) + H2O(l) → H2SO4(aq) + HCl(g) c. BeO(s) + C(s) → Be2C(s) + CO2(g)

Solution Steps

For the reaction:
\[ \mathrm{NaO}_{2}(s) + \mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{NaOH}(aq) + \mathrm{O}_{2}(g) + \mathrm{H}_{2} \mathrm{O}_{2}(aq) \]

1. Balance Sodium (Na):

   • 1 Na on both sides.

2. Balance Oxygen (O):

   • Left: 2 O from NaO\(_2\) + 1 O from H\(_2\)O = 3 O
   • Right: 1 O from NaOH + 2 O from O\(_2\) + 2 O from H\(_2\)O\(_2\) = 5 O
   • Adjust by adding 2 H\(_2\)O on the left:\[ \mathrm{NaO}_{2}(s) + 2 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{NaOH}(aq) + \mathrm{O}_{2}(g) + \mathrm{H}_{2} \mathrm{O}_{2}(aq) \]

3. Balance Hydrogen (H):

   • Left: 4 H from 2 H\(_2\)O
   • Right: 1 H from NaOH + 2 H from H\(_2\)O\(_2\) = 3 H
   • Adjust by adding 2 NaOH on the right:
     \[ \mathrm{NaO}_{2}(s) + 2 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow 2 \mathrm{NaOH}(aq) + \mathrm{O}_{2}(g) + \mathrm{H}_{2} \mathrm{O}_{2}(aq) \]

For the reaction:
\[ \mathrm{Fe}_{2} \mathrm{O}_{3}(s) + \mathrm{HBr}(aq) \rightarrow \mathrm{FeBr}_{3}(aq) + \mathrm{H}_{2} \mathrm{O}(l) \]

1. Balance Iron (Fe):

   • 2 Fe on the left, so 2 FeBr\(_3\) on the right.

2. Balance Bromine (Br):

   • Right: 6 Br from 2 FeBr\(_3\)
   • Left: 6 HBr needed.

3. Balance Oxygen (O):

   • 3 O from Fe\(_2\)O\(_3\) on the left, so 3 H\(_2\)O on the right.

4. Balance Hydrogen (H):

   • Left: 6 H from 6 HBr
   • Right: 6 H from 3 H\(_2\)O

Final balanced equation:
\[ \mathrm{Fe}_{2} \mathrm{O}_{3}(s) + 6 \mathrm{HBr}(aq) \rightarrow 2 \mathrm{FeBr}_{3}(aq) + 3 \mathrm{H}_{2} \mathrm{O}(l) \]

For the reaction:
\[ \mathrm{H}_{2} \mathrm{~S}(g) + \mathrm{O}_{2}(g) \rightarrow \mathrm{SO}_{2}(g) + \mathrm{H}_{2} \mathrm{O}(g) \]

1. Balance Sulfur (S):

   • 1 S on both sides.

2. Balance Oxygen (O):

   • Right: 2 O from SO\(_2\) + 1 O from H\(_2\)O = 3 O
   • Left: 1.5 O\(_2\) needed, but use 2 O\(_2\) to avoid fractions.

3. Balance Hydrogen (H):

   • Left: 2 H from H\(_2\)S
   • Right: 2 H from H\(_2\)O

Final balanced equation:
\[ 2 \mathrm{H}_{2} \mathrm{~S}(g) + 3 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{SO}_{2}(g) + 2 \mathrm{H}_{2} \mathrm{O}(g) \]

Final Answer