When `22.4 L` of `H_(2)(g)` is mixed with 11.2 of `Cl_(2)(g)`, each at STP, the moles of HCl(g) formed is equal to

Options

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)

Answer

A

Explanation

22.4 L of `H_(2)` at S.T.P =1 mol
11.2 L of `Cl_(2)` at STP = 0.5 mol
So, `Cl_(2)` is limiting reagent.
`H_(2) + Cl_(2) to 2HCl`
1 mol of `Cl_(2)` reacts with 1 mol of `H_(2)` to give 2 mol of HCl
`therefore 0.5` mol of `Cl_(2)` will react with 0.5 mol of `Cl_(2)` to give 1 mol of HCl.