Question

How many grams of excess reactant remain when 44.7 g of nitrogen gas and 57.3 g of oxygen gas are bubbled into excess water according to the equation below?

2 N2(g) + 5 O2(g) + 2 H2O(l) 4 HNO3(aq)

Answer 1

To find out how many grams of excess reactant remain, we need to determine the limiting reactant by calculating the moles of nitrogen gas and oxygen gas. Then, we compare the moles to the stoichiometry of the balanced equation. Finally, we use the moles of the excess reactant consumed to calculate the remaining grams.

Step-by-step explanation:

To determine how many grams of excess reactant remain, we need to find the limiting reactant first. Let's use the given amounts of nitrogen gas (N2) and oxygen gas (O2) to calculate how many moles of each are present. Then, we can compare the moles to the stoichiometry of the balanced equation to identify the limiting reactant.

Moles of N2 = 44.7 g / molar mass of N2
Moles of O2 = 57.3 g / molar mass of O2

Next, we determine the stoichiometric ratio between N2 and water (H2O) in the balanced equation and calculate the moles of excess reactant used up in the reaction.

Finally, we use the moles of the excess reactant consumed to calculate the grams of the excess reactant that remain.