De dot chay het 22g hon hop gom khi CH4 va H2 can 78,4l oxi (dktc)
a)Tinh thanh phan % ve moi khi trong hon hop
b)Tinh khoi luong nuoc thu duoc
Bai lam
Goi so mol cua khi CH4 la x(mol)
Goi so mol cua khi H2 la y (mol)
So mol cua 78,4l O2 la ;
nO2= 78,4/22,4= 3,5 ( mol)
PTHH; (1) CH4 + 2O2 ------> CO2 + 2H2O
1 mol 2 mol
x mol 2x mol
(2) 2H2 + O2 ------> 2H2O
2 mol 1 mol
y mol 1/2y mol

Tu phuong trinh (1) va (2), ta co pt:
nO2 = 2x + 1/2y = 3,5 mol (3)
Mat khac ta lai co:
mhh = mCH4 + mH2 = 16x + 2y =22 g ( 4 )
Ket hop (3) va ( 4 ) , giai ra ta duoc
x =1 mol va y= 3 mol
=> mCH4 = 1 * 16 =16 g
=> mH2 = 3 * 2 = 6 g
a) Vay thanh phan % cua hai khi trong hon hop la:
% CH4 = 16/22 *100 =72,7%
% H2 = 100% - 72,73 % = 27,3%
b) Theo PTHH 1: nH2O = 2*nCH4 = 2* 1 = 2 mol
Theo PTHH 2 : nH2O = nH2 = 3 mol




Tong so mol cua H2O trong ca 2 pt la;
3 + 2 = 5 mol
=> mH2O = 5 * 18 = 90 g