2H2+O2-to>2H2O
0,3---0,15------0,3
n H2=1 mol
n O2=0,15 mol
=>H2 dư=X
m H2O=0,3.18=5,4g
\(n_{H_2}=\dfrac{2}{2}=1mol\)
\(n_{O_2}=\dfrac{3,36}{22,4}=0,15mol\)
\(2H_2+O_2\rightarrow\left(t^o\right)2H_2O\)
1 < 0,15 ( mol )
0,15 0,3 ( mol )
Khí X là H2
\(m_{H_2O}=0,3.18=5,4g\)