Question

Answer 1

\(n_{H_2O\left(b\text{đ}\right)}=\dfrac{36}{18}=2\left(mol\right)\\ m_{H_2O\left(p\text{ư}\right)}=\dfrac{2.90}{100}=1,8\left(mol\right)\\ pthh:2H_2O\underrightarrow{t^o}2H_2+O_2\)
           1,8                   0,9 
\(V_{O_2}=0,9.22,4=20,16\left(l\right)\) 
\(n_{H_2O\left(\text{chưa điện phân}\right)}=2-1,8=0,2\left(mol\right)\\ m_{H_2O\left(\text{chưa điện phân}\right)}=0,2.18=3,6\left(g\right)\)