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?

Answers (1)

\frac{1}{2}H_2 +\frac{1}{2}Cl_2 \rightarrow HCl

moles of H_2 available =1

moles of Cl_2 available=\frac{1}{2}

Moles of HCl formed are stoichiometry will be 1.

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