\(n_{CH4}=\frac{1,6}{16}=0,1\left(mol\right)\)
\(CH_4+2O_2\underrightarrow{^{to}}CO_2+2H_2O\)
0,1____0,2_____________________(mol)
Theo phương trình ta có:
\(n_{O2}=2n_{CH4}=2.0,1=0,2\left(mol\right)\)
\(m_{O2}=0,2.32=6,4\left(g\right)\)
nCH4=1,6/16=0,1
CH4+2O2->CO2+2H20
1 : 2 : 1 : 2
0,1 0,2 0,1 0.2 (mol)
mO2=0,2*32=6,4(g)