volume problem?

Question

the reaction: 7H2O2 (aq) + N2H4 (g) ---> 2HNO3 (g) + 8H2O (g)



If a chemist starts with 110.0 grams of H2O2 and an excess of N2H4, what volume of water vapor will be produced at a temperature of 341oC and a pressure of 2.1 atm?

Answer 1

Which planet is the chemist now passing? First,convert 110 g H2O2 to moles .
Multiply moles of H2O2 by 8/7 (see equation)
You now have moles of H2O, so you can do the pV=nRT thing to get your answer of volume.

Answer 2

110.0 g H2O2 /(34.02 g/mol) = 3.233 mol H2O2
3.233 mol (8 mol H2O / 7 mol H2O2) = 3.695 mol H2O
PV=nRT
V=nRT/P
V= (3.695)(.0821)(341+273)/(2.1) = 88.7 L H2O