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 -

When 22.4 litres of H_(2(g))is mixed with 11.2 litres of Cl_(2(g)),eac

1.	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 -
Answer» 1 MOLE of `HCl_((g))` 2 mole of `HCl_((g))` 0.5 mole of `HCl_((g))` 1.5 mole of `HCl_((g))` SOLUTION : `H_(2)+Cl_(2) rarr 2HCL` `22.4 "lt."11.2 "lt."` 1 moleof `HCl=(1)/(2)` mole Limiting REAGENT is `Cl_(2)`. So, 1 mole HCl is formed.