Question

In the following reaction, when a surplus of hydrazine (N2H4) reacts with 20 liters of hydrogen peroxide (H2O2), how many liters of nitrogen gas will be released, given that both gases are at STP?
N2H4 + 2H2O2 dN2 + 4H2O
A) 10 liters
B) 0.1 liters
C) 40 liters
D) 20 liters

Answer 1

@thomaster or @chmvijay. help please stoichiometry =/

Answer 2

Try write up the relative quantities you need for the reaction to run 1 mol times.

Answer 3

thank you, i got the awnswer wrong but I think I know what i did wrong @Frostbite

Answer 4

In that case explain to me what :)

Answer 5

If you like?

Answer 6

else in the future this is a relation that can help you if ever in doubt:
for a reaction a A + b B -> c C d D
It follows:
\[\frac{ n(A) }{ a }=\frac{ n(B) }{ b }=\frac{ n(C) }{ c }=\frac{ n(D) }{ d }\]
doing so we get:
\[\frac{ n(N _{2}) }{ 1 }=\frac{ n(H _{2}O _{2} )}{ 2 }\]
\[V(N _{2})=\frac{ V(H _{2}O _{2}) }{ 2 }\]