Question

If 3.95 g of N2H4 reacts and produces 0.750 L of N2, at 295 K and 1.00 atm, what is the percent yield of the reaction?

Answer 1

Equation:
N2H4 → N2 + 2H2

percentage yield = (experimental yield/theoretical yield) * 100%

experimental yield = 0.750 L = 1.0*10^-6 dm^3

Find the theoretical yield by finding the molar volume of N2 using:

molar volume of N2 = number of moles of N2 * 24 dm^3

Find the number of moles of N2 by finding the number of moles of N2H4 since it is 1:1 ratio:
number of moles of N2 = number of moles of N2H4 = (mass of N2H4)/(molecular mass of N2H4)

Then you can find the percentage yield.