) PTHH :
(1)2CO+O2−t0−>2CO2(1)2CO+O2−t0−>2CO2
(2)2H2+O2−t0−>2H2O(2)2H2+O2−t0−>2H2O
b) Theo−đề−bà−ta−có:nCO2=8,844=0,2(mol)Theo−đề−bà−ta−có:nCO2=8,844=0,2(mol)
Theo PTHH 1 ta có :
nO2=12nCO2=0,1(mol)nO2=12nCO2=0,1(mol)
nCO = nCo2 = 0,2 (mol)
=> nO2(2) = 9,6−0,1.3232=0,2(mol)9,6−0,1.3232=0,2(mol)
Theo PTHH 2 ta có :
nH2 = 2nO2=2.0,2=0,4(mol)
=> ⎧⎩⎨%mCO=0,2.280,2.28+0,4.2.100%=87,5%%mH2=100%−87,5%=12,5%
nếu đúng vote mình 5* nhá