Question #1eeff

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Question #1eeff

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Answer 1 · 2017-04-27T04:32:11

${7 mol H}_{2}$ $"7 mol H"_2"$ will produce $\approx 5 . {mol NH}_{3}$ $~~5color(white)(.)"mol NH"_3"$ .

Explanation:

Balanced Equation

$N_{2} + {3H}_{2}$ $"N"_2+"3H"_2$ $\to$ $rarr$ ${2NH}_{3}$ $"2NH"_3$

Since the amount of $N_{2}$ $"N"_2"$ was not stated, I am assuming that it is in excess, which means we don't need to worry about it.

All we have to do is multiply the given moles of $H_{2}$ $"H"_2"$ by the mole ratio between $H_{2}$ $"H"_2"$ and ${NH}_{3}$ $"NH"_3"$ in the balanced equation. Since we want to end up with ${NH}_{3}$ $"NH"_3"$ , we put it in the numerator.

$7color(red)cancel(color(black)("mol H"_2))xxoverbrace(2"mol NH"_3)^color(red)("mole ratio")/(3color(red)cancel(color(black)("mol H"_2)))="5 mol NH"_3"$ rounded to one significant figure

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Question #1eeff

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