A mixture of hydrogen peroxide, H2O2, and hydrazine, N2H4, can be used

as a rocket propellant. The reaction is:
7 H2O2(g) + N2H4(l) ® 2 HNO3(aq) + 8 H2O(l)

a) How many moles of H2O2 react with 0.477 mol N2H4? [1] ___________
b) How many grams of HNO3 can be produced in a reaction of 67.7 g
H2O2 with excess N2H4? [3]
c) How many grams of HNO3 can be produced in a reaction of 67.7 g
H2O2 with 10.00 mL of N2H4 ( d= 1.006 g/mL) if the yield of the reaction
is 76.5 %? [4]

A 3.34
b 35.95
c 27.4 g

1 answer

A is correct.
For B I obtained 35.8.
For C you have assumed (or calculated) that H2O2 is the limiting reagent. If that were correct, then your answer of 27.4 is ok; however, N2H4 is the limiting reagent and your answer is not correct. I calculate something like 15 g HNO3 but check my work. I just did a quickie calculation.