Question

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?

a. 25.9 g
b. 23.1 g
c. 51.9 g
d. 11.5 g
e. 50.4 g

Answer 1

To calculate the mass of oxygen needed to produce 49.0 g of NO₂, use the balanced equation and molar mass of NO₂ to calculate the theoretical yield, then set up a proportion to find the mass of oxygen needed.

Step-by-step explanation:

To find the mass of oxygen needed to produce 49.0 g of NO₂, we first need to calculate the theoretical yield of NO₂ using the balanced equation. From the equation, we can see that 1 mole of N₂H₄ reacts with 3 moles of O₂ to produce 2 moles of NO₂. Therefore, we can use the molar mass of NO₂ to calculate the theoretical yield:

2 mol NO₂ = 46.01 g NO₂

To calculate the mass of oxygen needed, we can set up a proportion:

(46.01 g NO₂) / (2 mol NO₂) = (49.0 g NO₂) / (x)

Solving for x, the mass of oxygen needed, we get x = 25.9 g. Therefore, option a - 25.9 g - is the correct answer.