HCl on oxidation gives water and chlorine. How many litres of chlorine on STP can be obtained starting with 36.5 gram of HCl.

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Solution

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$4 HCl + O_{2} \to 2 {Cl}_{2} + 2 H_{2} O 4 mol 2 mol 4 mol of HCl goves 2 mol of {Cl}_{2} Number of moles of HCl present = \frac{Mass}{Molar mass} = \frac{36.5}{36.5} = 1 mol So, 1 mol of HCl will give \frac{1}{4} \times 2 = 0.5 mol of {Cl}_{2} and, 1 mol of {Cl}_{2} at STP occupies 22.4 L So, 0.5 mol of {Cl}_{2} will occupy 22.4 \times 0.5 = 11.2 L Hence, 11.2 L of {Cl}_{2} can be obtained with 36.5 g of HCl .$

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