Question #53b23

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Question #53b23

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Answer 1 · 2017-09-19T00:32:35

${13 moles NH}_{3}$ $"13 moles NH"_3$

Explanation:

The idea here is that the coefficients added in the balanced chemical equation in front of each chemical species that takes part in the reaction tell you the mole ratios that exist between these chemical species.

In your case, you have

$N_{2 (g)} + 3 H_{2 (g)} \to 2 {NH}_{3 (g)}$ $"N"_ (2(g)) + 3"H"_ (2(g)) -> 2"NH"_ (3(g))$

As you can see, you have

$N_{2 (g)} \to no coefficient = a coefficient of 1$ $"N"_ (2(g)) -> "no coefficient = a coefficient of 1"$

$3 H_{2 (g)} \to a coefficient of 3$ $3"H"_ (2(g)) -> "a coefficient of 3"$

$2 {NH}_{3 (g)} \to a coefficient of 2$ $2"NH"_ (3(g)) -> "a coefficient of 2"$

This means that the for every $1$ $1$ mole of nitrogen gas that takes part in the reaction, the reaction consumes $3$ $3$ moles of hydrogen gas, i.e. you have a $1 : 3$ $1:3$ mole ratio between nitrogen gas and hydrogen gas, and produces $2$ $2$ moles of ammonia, i.e. you have a $1 : 2$ $1:2$ mole ratio between nitrogen gas and ammonia.

Now, the problem tells you that you have $19$ $19$ moles of hydrogen gas that react with enough nitrogen gas.

This means that the reaction will consume all the moles of hydrogen gas available and produce

$19 color(red)(cancel(color(black)("moles H"_2))) * overbrace("2 moles NH"_2/(3color(red)(cancel(color(black)("moles H"_2)))))^(color(blue)("from the balanced chemical equation")) = color(darkgreen)(ul(color(black)("13 moles NH"_3)))$

The answer is rounded to two sig figs.

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Question #53b23

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