Consider the gaseous reaction: N₂H₄(g)+ 3 O₂(g)→ 2 NO₂(g)+ 2 H₂O(g) If the above reaction has a percent yield of 98.5%,what mass in grams of oxygen is needed to produce 49.0 g of NO₂(g),assuming an excess of N₂H₄?

Question

Quizplus · Accepted Answer

Since there is an excess of N₂H₄, we can assume that all of the NO₂(g) produced came from the 3 O₂(g) we started with. Therefore, the 49.0 g of NO₂(g) produced is also the mass of NO₂(g) that reacted. Using the stoichiometry of the reaction, we can calculate the amount of oxygen needed to produce that amount of NO₂(g): 3 mol O₂(g) produces 2 mol NO₂(g) x mol O₂(g) produces 49.0 g NO₂(g) x = (49.0 g NO₂(g) x 3 mol O₂(g) x 1 mol O) / (2 mol NO₂(g) x molar mass of O) x = 51.9 g O Therefore, the mass of oxygen needed to produce 49.0 g of NO₂(g) with a percent yield of 98.5% is 51.9 g, which is answer choice B.

Consider the Gaseous Reaction