Question

3 N2H4 + 4 ClF3 → 12 HF + 3 N2 + 2 Cl2

How much HF is produced by the reaction above?

Answer 1

1) find the number of substances:

N2H4 

$n=\frac{m}{M}=\frac{40}{32.046}=1.247$

excess ClF3, solved to N2H4 

$\frac{1.247}{3}=\frac{x}{12}$

x=4.988

1 )find the mass of HF

$m=n\times M=4.988\times20.006=99.789$ or 100 gr