Chlorine gas is prepared by reaction of H 2 S O 4 with M n O 2 and NaCl. What volume of C l 2 will be produced at STP if 50 g of NaCl is taken in the reaction?

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2 N a C l 2 moles 2 × 58.5 = 117 g + M n O 2 + 3 H 2 S O 4 → 2 N a H S O 4 + M n S O 4 + C l 2 1 mole 22.4 L STP + 2 H 2 O 117 g of NaCl ≡ 22.4 L of C l 2 50 g of N a C l ≡ 22.4 117 × 50 = 9.57 L of C l 2 at STP.

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