Consider the reaction: 4 NH₃ + 5 O₂ → 6 H₂O + 4 NO
If 125.0 g of NH₃ reacts with 225.0 g of O₂, how many grams of H₂O will be present after a complete reaction?

Question

Consider the reaction: 4 NH₃ + 5 O₂ → 6 H₂O + 4 NO If 125.0 g of NH₃ reacts with 225.0 g of O₂, how many grams of H₂O will be present after a complete reaction?

Quizplus · Accepted Answer

First, we need to interpret the chemical equation correctly by removing the subscript formatting issues. The correct chemical equation is: $4 NH_3 + 5 O_2 \rightarrow 6 H_2O + 4 NO$. To find out how many grams of $H_2O$ will be produced, we first determine the limiting reactant. The molar mass of $NH_3$ is approximately 17 g/mol, so 125.0 g corresponds to about 7.35 moles. The molar mass of $O_2$ is approximately 32 g/mol, so 225.0 g corresponds to about 7.03 moles. According to the stoichiometry of the reaction, 4 moles of $NH_3$ react with 5 moles of $O_2$, so $O_2$ is the limiting reactant. Using the stoichiometry of the reaction, 5 moles of $O_2$ produce 6 moles of $H_2O$. Therefore, 7.03 moles of $O_2$ will produce $7.03 \times \frac{6}{5} = 8.436$ moles of $H_2O$. The molar mass of $H_2O$ is approximately 18 g/mol, so $8.436$ moles of $H_2O$ correspond to $8.436 \times 18 \approx 151.9$ grams of $H_2O$.

Consider the Reaction: 4 NH3 + 5 O2 → 6

Consider the Reaction: 4 NH₃ + 5 O₂ → 6