WEBVTT

1
00:00:00.000 --> 00:00:10.182
Here is the problem: In the reaction N2 plus H2 → NH3, how many grams of NH3 form from 28 g of N2?

2
00:00:10.632 --> 00:00:15.450
Balance atoms of every element, then use the coefficients as mole ratios.

3
00:00:16.250 --> 00:00:25.410
Write an atom balance for each element. Give each species an unknown coefficient; atoms of every element must be equal on both sides.

4
00:00:25.860 --> 00:00:35.520
Solve the linear system with the smallest whole numbers. Exact elimination leaves one free coefficient; scaling by the least common denominator gives integers.

5
00:00:35.969 --> 00:00:37.118
Balanced equation.

6
00:00:37.568 --> 00:00:38.694
Moles of N2.

7
00:00:39.494 --> 00:00:42.513
NH3 via the mole ratio 2: 1.

8
00:00:43.313 --> 00:00:57.860
So the answer is N2 plus 3H2 → 2NH3; 28 g N2 → NH3: 34.045 g (1.999 mol).

9
00:00:58.310 --> 00:01:15.168
Coefficients in a balanced equation count particles (moles), not grams. Convert grams to moles, apply the mole ratio, then convert back to grams with the other species prime molar mass. Assumes the reaction goes to completion with the known species fully consumed (or formed).

10
00:01:15.968 --> 00:01:20.995
N: 2 equals 2, H: 6 equals 6, which checks out.

11
00:01:21.445 --> 00:01:26.507
NH3 is produced in a 2: 1 mole ratio to N2.
