Question #1eeff

1 Answer
Meave60
Apr 27, 2017

#"7 mol H"_2"# will produce #~~5color(white)(.)"mol NH"_3"#.

Explanation:

Balanced Equation

#"N"_2+"3H"_2##rarr##"2NH"_3#

Since the amount of #"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"# by the mole ratio between #"H"_2"# and #"NH"_3"# in the balanced equation. Since we want to end up with #"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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