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When 22.4 litres of `H_(2)(g)` is mixied with 11.2 litres of `Cl_(2)(g)`, each at S.T.P, the moles of HCl (g) formed is equal to :A. 0.5 mol of HCl (g)B. 1.5 mol of HCl (g)C. 1.0 mol of HCl (g)D. 2.0 mol of HCl (g) |
Answer» Correct Answer - C No. of moles in 22.4 L of `H_(2)(g)` at S.T.P = 1 mol No. of moles in 11.2 L of `Cl_(2)(g)` at S.T.P = 0.5 mol `underset("1 mol")(H_(2)(g))+underset("1 mol")(Cl_(2)(g))rarrunderset("2 mol")(2HCl(g))` `Cl_(2)(g)` is the limiting reactant since it has only 0.5 mole available for chemical reaction `:.` No. of moles of HCl (g) formed = 1mol. |
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