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Q

quynh_221195

nN2=1\3nH2
Ban dau nH2=3 mol
nN2=1 mol
goi nN2 pu=x mol
N2 + 3H2 ----> 2NH3
1 3 0
x 3x 2x
1-x 3-3x 2x
28(1-x)+2(3-3x)+34x=9(4-2x)
x=1\9
H=11,11%
 
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