When 10.0 g of NH_3 reacts completely with excess O_2 according to the equation 4 NH_3 + 5 O_2 → 4 NO + 6 H_2O, how many grams of H_2O are produced?

• A. 18.0 g
• B. 14.1 g
• C. 6.0 g
• D. 8.8 g

Answer: (B)

Verified step by step guidance

1. Write down the balanced chemical equation: \$4 \mathrm{NH}_3 + 5 \mathrm{O}_2 \rightarrow 4 \mathrm{NO} + 6 \mathrm{H}_2\mathrm{O}$.
2. Calculate the molar mass of ammonia (\(\mathrm{NH}_3\)). Use atomic masses: N = 14.0 g/mol, H = 1.0 g/mol, so \(M_{\mathrm{NH}_3} = 14.0 + 3 \times 1.0 = 17.0\) g/mol.
3. Convert the given mass of ammonia (10.0 g) to moles using the formula: \(\text{moles of } \mathrm{NH}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{10.0}{17.0}\) mol.
4. Use the stoichiometric ratio from the balanced equation to find moles of water produced. According to the equation, 4 moles of \(\mathrm{NH}_3\) produce 6 moles of \(\mathrm{H}_2\mathrm{O}\). So, \(\text{moles of } \mathrm{H}_2\mathrm{O} = \left(\frac{6}{4}\right) \times \text{moles of } \mathrm{NH}_3\).
5. Calculate the mass of water produced by multiplying moles of water by the molar mass of water (\(M_{\mathrm{H}_2\mathrm{O}} = 18.0\) g/mol): \(\text{mass of } \mathrm{H}_2\mathrm{O} = \text{moles of } \mathrm{H}_2\mathrm{O} \times 18.0\) g/mol.