According to the reaction N2O3(g) + 6H2(g) ->2NH3(g) + 3H2O(g) how many moles of NH3(g) could be formed from the reaction of 0.22 mol of N2O3(g) with 0.87 moles of H2.
`N_2O_3(g) + 6H_2(g) ->2NH_3(g) + 3H_2O(g) `
`N_2O_3:H_2 = 1:6`
Amount of `N_2O_3` added `= 0.22mol`
Amount of `H_2` added `= 0.87mol`
Amount of `H_2` required to react with `0.22` moles of `N_2O_3` is `0.22xx6 = 1.32`
Though the mole ratio for `N_2O_3:H_2` of the reaction is 1:6, the added mole ratio is less than this. So all the `N_2O_3` will not react. Instead of that all the `H_2` will react.
Amount of `H_2` reacted `= 0.87mol`
`H_2:NH_3 = 6:2 = 3:1`
Amount of `NH_3` produced `= 0.87/3 = 0.29`
So 0.29 moles of `NH_3` will be produced.
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