For the formation of 3.65g of HCI gas , what volume of hydrogen gas and chlorine gas , are required at NTP conditions?

Options

A. 1L, 1L
B. 1.12 L, 2.24 L
C. 3.65 L,1.83 L
D. 1.12L,1.12 L

Answer

D

Explanation

`H_(2)+CI_(2) to 2HCI" "n _(HCI) (3.65)/(36.5) = 0.1 `
`{:(2gm,71 gm,73gm,),(1"mole",1"mole",2 "mole",):}`
here 2 mole HCI produced by 1 mole `H _(2)`
1 mole HCI produce by 0.05 mole
`n_(H2) = V _(CI2) = 0.05 xx 22.4 =1.12L`