Question

Answer 1

a) \(M_2\left(CO_3\right)_n+nH_2SO_4\rightarrow M_2\left(SO_4\right)_n+nCO_2+nH_2O\)

b) Giả sử có 100g dd H₂SO₄ 9,8%

\(n_{H_2SO_4}=\dfrac{100.9,8\%}{98}=0,1\left(mol\right)\)

PTHH: M₂(CO₃)_n + nH₂SO₄ --> M₂(SO₄)_n + nCO₂ + nH₂O

              \(\dfrac{0,1}{n}\)<--------0,1--------->\(\dfrac{0,1}{n}\)--------->0,1

\(m_{dd.sau.pư}=\dfrac{0,1}{n}\left(2.M_M+60n\right)+100-0,1.44=\dfrac{0,1}{n}\left(2.M_M+60n\right)+95,6\left(g\right)\)

\(m_{M_2\left(SO_4\right)_n}=\dfrac{0,1}{n}\left(2.M_M+96n\right)\left(g\right)\)

=> \(C\%_{muối}=\dfrac{\dfrac{0,1}{n}\left(2.M_M+96n\right)}{\dfrac{0,1}{n}\left(2.M_M+60n\right)+95,6}.100\%=14,18\%\)

=> M_M = 28n (g/mol)

Xét n = 2 thỏa mãn => M_M = 56 (g/mol)

=> M là Fe