Calculate the number of moles of H2O2 required to produce 0.08 moles of HNO3 by the following reaction. 7 H2O2 + N2H4 2 HNO3 + 8 H2O

im having trouble with stoichiometry

First step is always balance the equation:

N2H4 + 7 H2O2 - - > 2HNO3 + 8H2O

So assuming that N2H4 is in excess and only moles of H2O2 are determining the product content then 7 moles of H2O2 give 2 moles of HNO3

So:

7:2

X:0.08

X=0.28 moles of H2O2