Question

Question #c35b1

Answer 1

`"0.5 moles N"_2`
`"4 moles H"_2`

Explanation:

Start by taking a look at the balanced chemical equation for this reaction

`3"H"_text(2(g]) + "N"_text(2(g]) -> color(red)(2)"NH"_text(3(g])`

Now, you know that the reaction produced `4` moles of ammonia. Your strategy now is to use the mole ratios that exist between the species that take part in the reaction to determine how many moles of each reactant actually took part in the reaction.

So, if the reaction produced `4` moles of ammonia, and you have a `3:color(red)(2)` mole ratio between hydrogen and ammonia, it follows that the reaction must have used up

`4color(red)(cancel(color(black)("moles NH"_3))) * ("3 moles H"_2)/(color(red)(2)color(red)(cancel(color(black)("moles NH"_3)))) = "6 moles H"_2`

By the same logic, the `1:color(red)(2)` mole ratio that exists between nitrogen and ammonia tells you that the reaction must have consumed

`4color(red)(cancel(color(black)("moles NH"_3))) * ("1 mole N"_2)/(color(red)(2)color(red)(cancel(color(black)("moles NH"_3)))) = "2 moles N"_2`

To test the result, check to see if the number of moles of each reactant satisfy the `3:1` mole ratio that exists between hydrogen and nitrogen.

As you can see, `6` moles of hydrogen would indeed need `2` moles of nitrogen.

Since you started with `2.5` moles of nitrogen, it follows that you're left with

`n_(N_2) = 2.5 - 2 = color(green)("0.5 moles N"_2)`

Likewise, the number of moles of hydrogen that remain will be

`n_(H_2) = 10 - 6 = color(green)("4 moles H"_2)`