When 22.4 litres of `H_(2)(g)` is mixed with 11.2 litres 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

`1mol-=22.4` litres at `STP,n_(H_(2))=(22.4)/(22.4)=1` mol
`n_(Cl_(2))=(11.2)/(22.4)=0.5` mol
`{:(,H_(2)(g),+,Cl_(2)(g),to,2HCl(g)),("Initial:",1mol,,0.5mol,,0),("Final:",(1-0.5),,(0.5-0.5),,2xx0.5),(,=0.5" mol",,=0" mol",,=1" mol"):}`
`Cl_(2)` is limiting reagent. So, 1 mol of HCl (g) is formed.