To determine the mass of water produced by the reaction of 3.58 g of ammonia (NH3), we need to follow these steps:
1. **Write the balanced chemical equation for the reaction:**
\[
4 \text{NH}_3 + 5 \text{O}_2 \rightarrow 4 \text{NO} + 6 \text{H}_2\text{O}
\]
2. **Calculate the molar mass of NH3:**
- Nitrogen (N): 1 atom × 14.01 g/mol = 14.01 g/mol
- Hydrogen (H): 3 atoms × 1.008 g/mol = 3.024 g/mol
\[
\text{Molar mass of NH}_3 = 14.01 \text{g/mol} + 3.024 \text{g/mol} = 17.034 \text{g/mol}
\]
3. **Convert the mass of NH3 to moles:**
\[
\text{Moles of NH}_3 = \frac{3.58 \text{ g}}{17.034 \text{ g/mol}} = 0.210 \text{ mol}
\]
4. **Use the stoichiometric coefficients from the balanced equation to find the moles of water produced:**
According to the balanced equation, 4 moles of NH3 produce 6 moles of H2O. Therefore, the ratio is:
\[
\frac{6 \text{ moles H}_2\text{O}}{4 \text{ moles NH}_3} \rightarrow \text{For every mole of NH}_3, \, \frac{6}{4} = 1.5 \text{ moles H}_2\text{O}
\]
5. **Calculate the moles of H2O produced from the given moles of NH3:**
\[
\text{Moles of H}_2\text{O} = 0.210 \text{ mol NH}_3 \times 1.5 \text{ mol H}_2\text{O / mol NH}_3 = 0.315 \text{ mol H}_2\text{O}
\]
6. **Calculate the mass of H2O produced:**
The molar mass of H2O is:
- Hydrogen (H): 2 atoms × 1.008 g/mol = 2.016 g/mol
- Oxygen (O): 1 atom × 16.00 g/mol = 16.00 g/mol
\[
\text{Molar mass of H}_2\text{O} = 2.016 \text{g/mol} + 16.00 \text{g/mol} = 18.016 \text{g/mol}
\]
\[
\text{Mass of H}_2\text{O} = 0.315 \text{ mol} \times 18.016 \text{ g/mol} = 5.67 \text{ g}
\]
Therefore, the mass of water produced by the reaction of 3.58 g of ammonia is **5.67 g**. This answer is reported to three significant digits to match the precision of the given mass of ammonia.
Ammonia (NH3) chemically reacts with oxygen gas (O2) to produce nitric oxide (NO) and water (H2O).
What mass of water is produced by the reaction of 3.58 g of ammonia?
Be sure your answer has the correct number of significant digits.
1 answer
1 answer