For the reaction N2(g) + 2H2(g) → N2H4(l), if the percent yield for this reaction is 100.0%, what is the actual mass of hydrazine (N2H4) produced when 59.20 g of nitrogen reacts with 6.750 g of hydrogen? a. Molar mass of N2 = 28.01 g/mol
b. Molar mass of H2 = 2.016 g/mol
c. Molar mass of N2H4 = 32.05 g/mol

53.6 g of N₂H₄

Explanation:

The begining is in the reaction:

N₂(g) + 2H₂(g) → N₂H₄(l)

We determine the moles of each reactant:

59.20 g / 28.01 g/mol = 2.11 moles of nitrogen

6.750 g / 2.016 g/mol = 3.35 moles of H₂

1 mol of N₂ react to 2 moles of H₂

Our 2.11 moles of N₂ may react to (2.11 . 2) /1 = 4.22 moles of H₂, but we only have 3.35 moles. The hydrogen is the limiting reactant.

2 moles of H₂ produce at 100 % yield, 1 mol of hydrazine

Then, 3.35 moles, may produce (3.35 . 1)/2 = 1.67 moles of N₂H₄

Let's convert the moles to mass:

1.67 mol . 32.05 g/mol = 53.6 g

24g

Explanation: