Question

When 22.4 litres of H2 (g) is mixed with 11.2 (g)litres ofCl2 at 273 k at 1 atm the moles of HCl(g) formed​

Answer 1

At STP, 1 mole of any gas occupies a volume of 22.4 L.

At STP, 1 mole of any gas occupies a volume of 22.4 L.22.4lt

At STP, 1 mole of any gas occupies a volume of 22.4 L.22.4ltH

At STP, 1 mole of any gas occupies a volume of 22.4 L.22.4ltH 2

At STP, 1 mole of any gas occupies a volume of 22.4 L.22.4ltH 2

At STP, 1 mole of any gas occupies a volume of 22.4 L.22.4ltH 2

At STP, 1 mole of any gas occupies a volume of 22.4 L.22.4ltH 2

At STP, 1 mole of any gas occupies a volume of 22.4 L.22.4ltH 2 +

At STP, 1 mole of any gas occupies a volume of 22.4 L.22.4ltH 2 + 11.2lt

At STP, 1 mole of any gas occupies a volume of 22.4 L.22.4ltH 2 + 11.2ltCl

At STP, 1 mole of any gas occupies a volume of 22.4 L.22.4ltH 2 + 11.2ltCl 2

At STP, 1 mole of any gas occupies a volume of 22.4 L.22.4ltH 2 + 11.2ltCl 2

At STP, 1 mole of any gas occupies a volume of 22.4 L.22.4ltH 2 + 11.2ltCl 2

At STP, 1 mole of any gas occupies a volume of 22.4 L.22.4ltH 2 + 11.2ltCl 2

At STP, 1 mole of any gas occupies a volume of 22.4 L.22.4ltH 2 + 11.2ltCl 2 →2HCl

At STP, 1 mole of any gas occupies a volume of 22.4 L.22.4ltH 2 + 11.2ltCl 2 →2HClLimiting reagent is Cl

At STP, 1 mole of any gas occupies a volume of 22.4 L.22.4ltH 2 + 11.2ltCl 2 →2HClLimiting reagent is Cl 2

At STP, 1 mole of any gas occupies a volume of 22.4 L.22.4ltH 2 + 11.2ltCl 2 →2HClLimiting reagent is Cl 2

At STP, 1 mole of any gas occupies a volume of 22.4 L.22.4ltH 2 + 11.2ltCl 2 →2HClLimiting reagent is Cl 2 .

At STP, 1 mole of any gas occupies a volume of 22.4 L.22.4ltH 2 + 11.2ltCl 2 →2HClLimiting reagent is Cl 2 . 1 mole of chlorine forms 2 moles of HCl.

At STP, 1 mole of any gas occupies a volume of 22.4 L.22.4ltH 2 + 11.2ltCl 2 →2HClLimiting reagent is Cl 2 . 1 mole of chlorine forms 2 moles of HCl.Hence, 0.5 moles of chlorine will form 1 mole of HCl.

At STP, 1 mole of any gas occupies a volume of 22.4 L.22.4ltH 2 + 11.2ltCl 2 →2HClLimiting reagent is Cl 2 . 1 mole of chlorine forms 2 moles of HCl.Hence, 0.5 moles of chlorine will form 1 mole of HCl.Thus, when 22.4 liters of H

At STP, 1 mole of any gas occupies a volume of 22.4 L.22.4ltH 2 + 11.2ltCl 2 →2HClLimiting reagent is Cl 2 . 1 mole of chlorine forms 2 moles of HCl.Hence, 0.5 moles of chlorine will form 1 mole of HCl.Thus, when 22.4 liters of H 2

At STP, 1 mole of any gas occupies a volume of 22.4 L.22.4ltH 2 + 11.2ltCl 2 →2HClLimiting reagent is Cl 2 . 1 mole of chlorine forms 2 moles of HCl.Hence, 0.5 moles of chlorine will form 1 mole of HCl.Thus, when 22.4 liters of H 2

At STP, 1 mole of any gas occupies a volume of 22.4 L.22.4ltH 2 + 11.2ltCl 2 →2HClLimiting reagent is Cl 2 . 1 mole of chlorine forms 2 moles of HCl.Hence, 0.5 moles of chlorine will form 1 mole of HCl.Thus, when 22.4 liters of H 2 (g) is mixed with 11.2 liters of Cl

At STP, 1 mole of any gas occupies a volume of 22.4 L.22.4ltH 2 + 11.2ltCl 2 →2HClLimiting reagent is Cl 2 . 1 mole of chlorine forms 2 moles of HCl.Hence, 0.5 moles of chlorine will form 1 mole of HCl.Thus, when 22.4 liters of H 2 (g) is mixed with 11.2 liters of Cl 2

At STP, 1 mole of any gas occupies a volume of 22.4 L.22.4ltH 2 + 11.2ltCl 2 →2HClLimiting reagent is Cl 2 . 1 mole of chlorine forms 2 moles of HCl.Hence, 0.5 moles of chlorine will form 1 mole of HCl.Thus, when 22.4 liters of H 2 (g) is mixed with 11.2 liters of Cl 2

At STP, 1 mole of any gas occupies a volume of 22.4 L.22.4ltH 2 + 11.2ltCl 2 →2HClLimiting reagent is Cl 2 . 1 mole of chlorine forms 2 moles of HCl.Hence, 0.5 moles of chlorine will form 1 mole of HCl.Thus, when 22.4 liters of H 2 (g) is mixed with 11.2 liters of Cl 2 (g), each at STP, the moles of HCl(g) formed is equal to 1 mol of HCl(g).

At STP, 1 mole of any gas occupies a volume of 22.4 L.22.4ltH 2 + 11.2ltCl 2 →2HClLimiting reagent is Cl 2 . 1 mole of chlorine forms 2 moles of HCl.Hence, 0.5 moles of chlorine will form 1 mole of HCl.Thus, when 22.4 liters of H 2 (g) is mixed with 11.2 liters of Cl 2 (g), each at STP, the moles of HCl(g) formed is equal to 1 mol of HCl(g).

Explanation:

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