Question #71e84

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Question #71e84

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Answer 1 · 2017-07-27T14:27:05

$V_{N H_{3}} \approx 20 . L$ $V_(NH_3) approx 20. L$

Explanation:

This is a stoichiometry-gas law problem.

$N_{2} + 3 H_{2} \to 2 N H_{3}$ $N_2+3H_2 to 2NH_3$

$\frac{10 L}{22.4 L} = 0.446 m o l [N_{2}]$ $(10L)/(22.4L) = 0.446mol[N_2]$
$5 g \cdot \frac{2.02 g}{H_{2}} = 2.48 m o l [H_{2}]$ $5g* (2.02g)/H_2 = 2.48mol[H_2]$

Based on stoichiometry, you can see that $N_{2}$ $N_2$ will be the limiting reactant, so let's calculate the amount at STP.

$1.0 a t m \cdot 10.0 L = n \cdot \frac{0.08206 L \cdot a t m}{m o l \cdot K} \cdot 273.15 K = n = 0.446$ $1.0atm*10.0L = n*(0.08206L*atm)/(mol*K)*273.15K = n = 0.446$

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

$0.446 m o l \cdot \frac{2 N H_{3}}{N_{2}} \approx 0.892 m o l [N H_{3}] = n$ $0.446mol*(2NH_3)/(N_2) approx 0.892mol[NH_3] = n$

$1.0 \cdot V = 0.892 \cdot \frac{0.08206 L \cdot a t m}{m o l \cdot K} \cdot 273.15 K$ $1.0*V = 0.892*(0.08206L*atm)/(mol*K)*273.15K$
$thereforeV approx 20. L$

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Question #71e84

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