Câu 1 .
`n_(H_2) = 13.44 / 22.4 = 0.6 ( mol ) `
`BTNT ( H ) : n_(HCl) = 2n_(H_2) = 2× 0.6 = 1.2 ( mol ) `
`-> n_(Cl^-) = 1.2 ( mol ) `
`BTKH : m_(\text{muối}) = m_(\text{kim loại }) + m_(Cl^-)
`=> m_(\text{muối}) = 25.12 + 1.2 × 35.5 = 67.72 ( gam ) `
Câu 2 .
$\displaystyle\sum$`m_(Mg+Al) = 9.14 - 2.54 = 6.6 ( gam ) `
`n_(H_2) = 7.84/ 22.4 = 0.35 ( mol ) `
`BTNT ( H ) : n_(HCl) = 2n_(H_2) = 2× 0.35 = 0.7 ( mol ) `
` -> n_(Cl^-) = 1.2 ( mol ) `
`BTKH : m_(\text{muối}) = m_(\text{Mg+Al }) + m_(Cl^-) `
` => m_(\text{muối}) = 6.6 + 0.7 × 35.5 = 31.45 ( gam ) `
Câu 3 .
`n_(AgNO_3) = 0.2 × 0.1 = 0.2 ( mol ) `
`n_(FeCl_2) = 0.1 × 0.1 = 0.01 ( mol ) `
`PTHH : 2AgNO_3 + FeCl_2 -> 2AgCl ↓ + Fe(NO_3)_2`
` 0.02 ← 0.01 → 0.02 → 0.02`
` AgNO_3 + Fe(NO_3)_2 -> Fe(NO_3)_3 + Ag ↓`
` 0.02 ← 0.02 → 0.02`
$\displaystyle\sum$ `\text{kết tủa} = m_(AgCl) + m_(Ag) `
`=> m_(\text{kết tủa}) = 0.02 × 143.5 + 0.02 × 108 = 5.03 ( g ) `
