\(2X+nCl_2 \xrightarrow{t^{o}} 2XCl_n\\ n_{X}=a(mol)\\ n_{XCl_n}=a(mol)\\ \frac{a.X}{a.(X+35,5n)}=\frac{1,625}{3,4}\\ \to \frac{X}{X+35,5n}=\frac{1,625}{3,4}\\ \to X=32,5n(g/mol)\\ a=2; X=65 (Zn) \)

\(Mg+2HCl \to MgCl_2+H_2\\ n_{H_2}=0,15(mol)\\ \to n_{Mg}=n_{H_2}=0,15(mol)\\ \%m_{Mg}=\frac{0,15.24}{10}.100\%=36\%\\ \%m_{Cu}=100\%-36\%=64\%\)

PTHH: CaCO₃ ---t^o---> CaO + CO₂

Ta có: \(n_{CaCO_3}=\dfrac{50}{100}=0,5\left(mol\right)\)

Theo PT: \(n_{CaO}=n_{CaCO_3}=0,5\left(mol\right)\)

Vì CaO là chất răn nên thể tích bằng số mol của chính nó.

=> \(V_{CaO}=0,5\left(lít\right)\)

Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

a. PTHH: Fe + H₂SO₄ ---> FeSO₄ + H₂

b. Theo PT: \(n_{H_2}=n_{Fe}=0,1\left(mol\right)\)

=> \(V_{H_2}=0,1.22,4=2,24\left(lít\right)\)

c. Theo PT: \(n_{H_2SO_4}=n_{Fe}=0,1\left(mol\right)\)

=> \(m_{H_2SO_4}=0,1.98=9,8\left(g\right)\)

=> \(C_{\%_{H_2SO_4}}=\dfrac{9,8}{100}.100\%=9,8\%\)

PTHH: 2Al + 3H₂SO₄ ---> Al₂(SO₄)₃ + 3H₂.

Ta có: \(C_{\%_{H_2SO_4}}=\dfrac{m_{H_2SO_4}}{200}.100\%=14,7\%\)

=> \(m_{H_2SO_4}=29,4\left(g\right)\)

=> \(n_{H_2SO_4}=\dfrac{29,4}{98}=0,3\left(mol\right)\)

Theo PT: \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}.n_{H_2SO_4}=\dfrac{1}{3}.0,3=0,1\left(mol\right)\)

=> \(m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2\left(g\right)\)

Đổi 112ml = 0,112 lít, 700ml = 0,7 lít.

Ta có: \(n_{SO_2}=\dfrac{0,112}{22,4}=0,005\left(mol\right)\)

a. PTHH: SO₂ + Ca(OH)₂ ---> CaSO₃↓ + H₂O

Theo PT: \(n_{CaSO_3}=n_{SO_2}=0,005\left(mol\right)\)

=> \(m_{CaSO_3}=0,005.120=0,6\left(g\right)\)

b. Theo PT: \(n_{Ca\left(OH\right)_2}=n_{SO_2}=0,005\left(mol\right)\)

=> \(C_{M_{Ca\left(OH\right)_2}}=\dfrac{n_{ct}}{V_{dd}}=\dfrac{0,005}{0,7}\approx0,007\left(M\right)\)

\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

PTHH: Mg + H₂SO4 → MgSO₄ + H₂

Mol:      0,2                                    0,2

\(V_{H_2}=0,2.24,79=4,958\left(l\right)\)

 ⇒ Chọn C