\(n_{O_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\\
pthh:2H_2+O_2\underrightarrow{t^o}2H_2O\)
0,5 1
\(m_{H_2O}=1.18=18g\)
\(n_{O_2}=\dfrac{11,2}{22,4}=0,5mol\)
\(2H_2+O_2\rightarrow\left(t^o\right)2H_2O\)
0,5 1 ( mol )
\(m_{H_2O}=1.18=18g\)