Chlorine gas is prepared by reaction of H2SO4 with MnO2 and NaCl. What volume of Cl2 will be produced at STP if 50 g of NaCl is taken in the reaction?

Question

Chlorine gas is prepared by reaction of H2SO4 with MnO2 and NaCl. What volume of Cl2 will be produced at STP if 50 g of NaCl is taken in the reaction? (A) 1.915 L (B) 22.4 L (C) 11.2 L (D) 9.57 L

Tardigrade Admin · Accepted Answer

underset text2 moles(2× 58.5 = 117 g)2NaCl +MnO2+3H2SO4 → 2NaHSO4 +MnSO4 + underset text1 mole 22.4 L(STP)Cl2 +H2O 117 g of NaCl ≡ 22.4 L of Cl2 50 g of NaCl ≡ (22.4/117) × 50 = 9.57 L of Cl2 at STP

Q. Chlorine gas is prepared by reaction of $H_{2} S O_{4}$ with $M n O_{2}$ and $N a Cl$ . What volume of $C l_{2}$ will be produced at $STP$ if $50 g$ of $N a Cl$ is taken in the reaction?

$1.915 L$

$22.4 L$

$11.2 L$

$9.57 L$

$2 moles(2 \times 58.5 = 117 g) 2 N a Cl$ $+ M n O_{2}$ $+ 3 H_{2} S O_{4} \to 2 N a H S O_{4} + M n S O_{4} +$ $1 mole 22.4 L(STP) C l_{2}$ $+ H_{2} O$
$117 g$ of $N a Cl \equiv 22.4 L$ of $C l_{2}$
$50 g$ of $N a Cl \equiv \frac{22.4}{117} \times 50 = 9.57 L$ of $C l_{2}$ at $STP$

Q. Chlorine gas is prepared by reaction of H2​SO4​ with MnO2​ and NaCl. What volume of Cl2​ will be produced at STP if 50g of NaCl is taken in the reaction?

1.915L

22.4L

11.2L

9.57L