3 N2H4 + 4 ClF3 → 12 HF + 3 N2 + 2 Cl2
How much HF is produced by the reaction above?
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
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