Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with aqueous hydrochloric acid according to the reaction, 4HCl (aq) + MnO2(s)→ 2H2O(l) + MnCl2(aq) + Cl2(g) How many grams of HCl react with 5.0 g of manganese dioxide?

Question

Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with aqueous hydrochloric acid according to the reaction, 4HCl (aq) + MnO2(s)→ 2H2O(l) + MnCl2(aq) + Cl2(g) How many grams of HCl react with 5.0 g of manganese dioxide? (A) 8.4 g (B) 5.5 g (C) 3.4 g (D) 4.8 g

Tardigrade Admin · Accepted Answer

underset overset4 molor 146 g4HCl(aq) + underset overset1 molor 87 gMnO2(s) → 2 H2O(l) + MnCl2(aq) + Cl2(g) 87 g of MnO2 reacts with 146 g of HCl. ∴ 5 g of MnO2 will react with (146 × 5/87) = 8.39 ≈ 8. 40 g of HCl

Q. Chlorine is prepared in the laboratory by treating manganese dioxide $(M n O_{2})$ with aqueous hydrochloric acid according to the reaction,
$4 H C l_{(a q)} + M n O_{2 (s)} \to 2 H_{2} O_{(l)} + M n C l_{2 (a q)} + C l_{2 (g)}$
How many grams of $H Cl$ react with $5.0 g$ of manganese dioxide?

8.4 g

5.5 g

3.4 g

4.8 g

$or 146 g 4 m o l 4 H C l_{(a q)} + or 87 g 1 m o l M n O_{2 (s)} \to 2 H_{2} O_{(l)} + M n C l_{2 (a q)} + C l_{2 (g)}$
$87 g$ of $M n O_{2}$ reacts with $146 g$ of $H Cl$ .
$∴ 5 g$ of $M n O_{2}$ will react with
$\frac{146 \times 5}{87} = 8.39 \approx 8.40 g$ of $H Cl$

Q. Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2​) with aqueous hydrochloric acid according to the reaction, 4HCl(aq)​+MnO2(s)​→2H2​O(l)​+MnCl2(aq)​+Cl2(g)​ How many grams of HCl react with 5.0g of manganese dioxide?