Question #e811d

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Question #e811d

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Answer 1 · 2017-03-03T02:02:07

${0.43 moles N}_{2}$ $"0.43 moles N"_2$

Explanation:

Start by writing a balanced chemical equation that describes this synthesis reaction

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

The thing to remember about balanced chemical equations is that the coefficients added in front of a chemical species represent the number of moles of said chemical species needed in order for the reaction to take place.

In this case, you need $1$ $1$ mole of hydrogen gas for every $3$ $color(blue)(3)$ moles of hydrogen gas in order to produce $2$ $color(red)(2)$ moles of ammonia.

Now, these mole ratios are true regardless of how many moles of a particular chemical species you have. In other words, you will have

$\frac{{number of moles of N}_{2}}{{number of moles of NH}_{3}} = \frac{1}{2}$ $"number of moles of N"_2/"number of moles of NH"_3 = 1/color(red)(2)$

Similarly, you will also have

$\frac{{number of moles of H}_{2}}{{number of moles of N}_{2}} = \frac{3}{1}$ $"number of moles of H"_2/"number of moles of N"_2 = color(blue)(3)/1$

and

$\frac{{number of moles of H}_{2}}{{number of moles of NH}_{3}} = \frac{3}{2}$ $"number of moles of H"_2/"number of moles of NH"_3 = color(blue)(3)/color(red)(2)$

In your case, you know that $0.85$ $0.85$ moles of ammonia were produced by the reaction. Since the $1 : 2$ $1:color(red)(2)$ mole ratio that exists between nitrogen gas and ammonia tells you that

${no. of moles of N}_{2} = \frac{1}{2} \cdot {no. of moles of NH}_{3}$ $"no. of moles of N"_2 = 1/color(red)(2) * "no. of moles of NH"_3$

you can say that the reaction consumed

$"no. of moles of N"_2 = 1/color(red)(2) * "0.85 moles" = color(darkgreen)(ul(color(black)("0.43 moles N"_2)))$

The answer must be rounded to two sig figs, the number of sig figs you have for the number of moles of ammonia produced by the reaction.

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Question #e811d

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