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

To determine the mass of oxygen needed, we first need to use stoichiometry to find the moles of NO₂(g) produced: 1 mol N₂H₄(g) + 3 mol O₂(g) → 2 mol NO₂(g) + 2 mol H₂O(g) From the balanced equation, we can see that 3 moles of oxygen are needed to produce 2 moles of NO₂(g). Therefore, to find the moles of oxygen needed to produce 49.0 g of NO₂(g), we can use the molar mass of NO₂(g): 49.0 g NO₂(g) * (1 mol NO₂(g) / 92.06 g NO₂(g)) * (3 mol O / 2 mol NO₂(g)) = 1.889 mol O Next, we need to convert moles of oxygen to grams: 1.889 mol O * 32.00 g/mol = 60.44 g O However, we know that the percent yield of the reaction is 98.5%, so the actual yield will be: Actual yield = 98.5% * Theoretical yield Actual yield = 0.985 * 60.44 g O Actual yield = 59.51 g O Therefore, the mass in grams of oxygen needed is approximately 59.51 g, which is closest to choice B (51.9 g).

Consider the Gaseous Reaction: N2H4(g) + 3 O2(g) → 2

Consider the Gaseous Reaction: N₂H₄(g) + 3 O₂(g) → 2