`a)`

Bs: tỉ lệ thể tích tương ứng là 1:2

Giả sử `n_{Cl_2}=1(mol)->n_{O_2}=2(mol)`

`\%V_{Cl_2}=1/{1+2}.100\%=33,33\%`

`\%V_{O_2}=100-33,33=66,67\%`

`\%m_{Cl_2}={1.71}/{1.71+2.32}.100\%=52,59\%`

`\%m_{O_2}=100-52,59=47,41\%`

`b)`

`M_A={1.71+2.32}/{1+2}={103}/3(g//mol)`

`->d_{A//H_2}={{103}/3}/2=17,17`

`c)`

`n_A={7,437}/{24,79}=0,3(mol)`

`->m_A=0,3.{103}/3=10,3(g)`