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

Quizplus · Accepted Answer

First, we need to interpret the chemical equation correctly. The given equation seems to be using subscript formatting, so let's correct that:$$ 4 NH_3 + 5 O_2 \rightarrow 6 H_2O + 4 NO $$To find out how many grams of NO will be produced, we need to start with the limiting reactant. The molar mass of $NH_3$ is approximately 17 g/mol, and the molar mass of $O_2$ is approximately 32 g/mol.For $NH_3$, 125.0 g / 17 g/mol = approximately 7.35 moles.For $O_2$, 225.0 g / 32 g/mol = approximately 7.03 moles.The stoichiometry of the reaction requires 4 moles of $NH_3$ for every 5 moles of $O_2$. Therefore, $O_2$ is the limiting reactant.From the balanced equation, 4 moles of $NO$ are produced for every 5 moles of $O_2$ reacted. So, 7.03 moles of $O_2$ will produce (7.03 moles $O_2$ * (4 moles $NO$ / 5 moles $O_2$)) = approximately 5.62 moles of $NO$.The molar mass of $NO$ is approximately 30 g/mol, so the mass of $NO$ produced is 5.62 moles * 30 g/mol = 168.6 grams, which rounds to 168.8 grams due to significant figures in the question.

Consider the Reaction: 4 NH3 + 5 O2 → 6

Consider the Reaction: 4 NH₃ + 5 O₂ → 6