Question

Question #d83e7

Answer 1

Here's what I got.

Explanation:

Since you can't really have `3.01` molecules of nitrogen gas, I'll assume that you're actually dealing with `3.01 * 10^(23)` molecules of nitrogen gas, the equivalent of

`3.01 * 10^(23)color(red)(cancel(color(black)("molecules N"_2))) * overbrace("1 mole N"_2/(6.02 * 10^(23)color(red)(cancel(color(black)("molecules N"_2)))))^(color(blue)("Avogadro's constant"))`

`= "0.500 moles N"_2`

Now, nitrogen gas and hydrogen gas react to form ammonia according to the following balanced chemical equation

`"N"_ (2(g)) + 3"H"_ (2(g)) -> 2"NH"_ (3(g))`

Notice that it takes `1` mole of nitrogen gas to produce `2` moles of ammonia, which means that your reaction will produce--keep in mind that the nitrogen gas reacts completely!

`0.500 color(red)(cancel(color(black)("moles N"_2))) * "2 moles NH"_3/(1color(red)(cancel(color(black)("mole N"_2)))) = "1.00 moles NH"_3`

Under Standard Conditions for Pressure and Temperature, `1` mole of any ideal gas occupies `"22.7 L"` `->` this is known as the molar volume of a gas at STP.

You can thus say that your reaction will produce

`1.00 color(red)(cancel(color(black)("moles NH"_3))) * overbrace("22.7 L"/(1color(red)(cancel(color(black)("mole NH"_3)))))^(color(blue)("under STP conditions")) = color(darkgreen)(ul(color(black)("22.7 L")))`

The answer is rounded to three sig figs, the number of significant figures you have for the number of molecules of nitrogen gas.