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)
H2 + 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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