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When 22.4 litres of H2(g) is mixed with 11.2 litres of Cl2(g), each at S.T.P, the moles of HCl(g) formed is equal to

(a) 1 mol of HCl(g)

(b) 2 mol of HCl(g)

(c) 0.5 mol of HCl(g)

(d) 1.5 mol of HCl(g).

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Answer - (a) 1 mol of HCl(g)

 

Solution - 

H+ Cl2    ⇒ 2HCl 

no. of moles of H2  = 22.4/22.4 = 1 mole 

no. of moles of Cl2 = 11.2/22.4 = 0.5 mole

thus limiting reagent is Cl2 

for 1 mole Cl2 → 2 mole HCl

for 0.5 mole Cl2 → 1 mole HCl

Answer = 1 mole HCl

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