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)` = 1 mol of `H_(2)`
11.2 L of `Cl_(2)=(11.2)/(22.4)"mol = 0.5 mol of "Cl_(2)`
`underset("1 mol")(H_(2)(g))+underset("1 mol")(Cl_(2)(g))rarrunderset("2 mol")(2HCl)(g)`
Thus, `Cl_(2)` will be the limiting reagent and HCl formed by reaction of 0.5 mol of `Cl_(2)=1` mol