\(a)m_{O_2}=20,3-13,1=7,2g\\ n_{O_2}=\dfrac{7,2}{32}=0,225mol\\ BTNT\left(O\right):n_{H_2O}=2n_{O_2}=0,225.2=0,45mol\\ BTNT\left(H\right):n_{HCl}=2n_{H_2O}=0,45.2=0,9mol\\ V_{HCl}=\dfrac{0,9}{0,4}=2,25l\\ b)BTKL:m_{oxit}+m_{HCl}=m_{muối}+m_{nước}\\ \Leftrightarrow20,3+0,9.36,5=m_{muối}+0,45.18\\ \Rightarrow m_{muối}=45,05g\)
P/s: giống bài lớp 9 vậy:)