# 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) `

Mole ratio

`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,...

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`N_2O_3(g) + 6H_2(g) ->2NH_3(g) + 3H_2O(g) `

Mole ratio

`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`

Mole ratio

`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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