Question

Question #71e84

Answer 1

`V_(NH_3) approx 20. L`

Explanation:

This is a stoichiometry-gas law problem.

`N_2+3H_2 to 2NH_3`

`(10L)/(22.4L) = 0.446mol[N_2]`
`5g* (2.02g)/H_2 = 2.48mol[H_2]`

Based on stoichiometry, you can see that `N_2` will be the limiting reactant, so let's calculate the amount at STP.

`1.0atm*10.0L = n*(0.08206L*atm)/(mol*K)*273.15K = n = 0.446`

Well, that was a waste of time, but it's nice to double check that they are the same, anyways:

`0.446mol*(2NH_3)/(N_2) approx 0.892mol[NH_3] = n`

`1.0*V = 0.892*(0.08206L*atm)/(mol*K)*273.15K`
`thereforeV approx 20. L`