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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:

  • Option 1)

    1 mol of HC(g)

  • Option 2)

    2 mol of HCl (g)

  • Option 3)

    0.5 mol of HC(g)

  • Option 4)

    1.5 mol of HCl (g)

 

Answers (1)

best_answer

\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.


Option 1)

1 mol of HC(g)

Correct

Option 2)

2 mol of HCl (g)

Incorrect

Option 3)

0.5 mol of HC(g)

Incorrect

Option 4)

1.5 mol of HCl (g)

Incorrect

Posted by

Aadil

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