`n_{H_2}=\frac{3,36}{22,4}=0,15(mol)`
Bảo toàn `H` `=>` `n_{H_2}=n_{H_2SO_4}=0,15(mol)`
`m_{\text{dd} H_2SO_4}=\frac{0,15.98.100%}{10%}=147g`
Bảo toàn khối lượng:
`m_{\text{dd} Y}=m_{\text{dd} H_2SO_4}+m_{\text{hh kim loại}}-m_{H_2}`
`=> m_{\text{dd}Y}=5,2+147-0,15.2=151,9g`