Câu 1 :
` n_(H_2) = 0.672 / 22.4 = 0.03 ( mol ) `
`PT : M + 2H_2O -> M(OH)_2 + H_2`
` 0.03 - 0.03 - 0.03 (mol ) `
`M_M = m_m / n_m = 1.2 / 0.03 = 40 ( đvC ) `
=> `M` là Canxi `( Ca ) `
`* m_(ddspu) = 1.2 + 100 - 0.03 * 2 = 101.14 ( g ) `
`C%Ca(OH)_2 =` $\frac{0.03 * 74}{101.14}$ `* 100% = 2.19 %`
Câu 2 /
Ta có : Δm↑ `= m_(kim loại) - m_(H_2) `
=> ` 1.32 = 1.38 - m_(H_2) `
=> `m_(H_2) = 0.06 ( g ) `
=> `n_(H_2) = 0.06 / 2 = 0.03 ( mol )`
`PT : 2D + 2HCl -> 2MCl + H_2`
` 0.06 - 0.06 - 0.03`
`M_D = m_D / n_D = 1.38 / 0.06 = 23 ( đvC )`
=> `D` là Natri `( Na )`
`m_(ddspu) = 1.38 + 80 - 0.03 * 2 = 81.32 ( g )`
`C%NaCl =` $\frac{0.06 * 58.5}{81.32}$ `* 100% = 4.32 %`
