`H_(2)(g) + I_(2)(g)hArr2HI(g)` When 46 g of `I_(2)` and 1 g of `H_(2)` gas heated at equilibrium at `450^(@)C`, the equilibrium mixture contained `1.9` g of `I_(2)` . How many moles of `I_(2)` and HI are present at equilibrium ?

Options

A. `0.0075` and `0.147` moles
B. `0.005` and `0.147` moles
C. `0.0075` and `0.347` moles
D. `0.0052` and `0.347` moles

Answer

C

Explanation

Moles of `I_(2)` taken `=(46)/(254) =0.181`
Moles of `H_(2)` taken `=(1)/(2) =0.5`
Moles of `I_(2)` remaining `=(1.9)/(254)=0.0075`
Mols of `I_(2)` used `=0.81 - 0.0075 = 0.1735`
Moles of `H_(2)` used `=0.1735`
Moles of `H_(2)` reamining `=0.5 - 0.1735 = 0.3265`
Moles of HI formed `=0.1735xx2=0.347`
At equilibrium, moles of `I_(2) =0.0075` moles
Moles of HI `=0.347` moles