Question

Question #65722

Answer 1

`"100 g"`

Explanation:

You are on the right track with this one.

Use the known ratio for the first mixture of `"A"`, `"B"`, and `"C"` to determine the mass of each chemical in the `"100-g"` sample.

If you take `m_1` to be the mass of chemical `"A"`, you can say that

`3 xx m_1 ->` the mass of chemical `"B"`

`1 xx m_1 ->` the mass of chemical `"C"`

This means that you have

`m_1 + 3 * m_1 + m_1 = "100 g"`

`5 * m_1 = "100 g" implies m_1 = "100 g"/5 = "20 g"`

Do the same for the second `"100-g"` mixture of `"A"`, `"B"`, and `"C"`. Let's say that `m_2` is the mass of chemical `"A"` in this mixture. You will have

`2 xx m_2 ->` the mass of chemical `"B"`

`7 xx m_2 ->` the mass of chemical `"C"`

This means that you have

`m_2 + 2 * m_2 + 7 * m_2 = "100 g"`

`10 * m_2 = "100 g" implies m_2 = "10 g"`

So, you're adding these two mixture together. The masses of the three chemicals will be

`m_A = "20 g" + "10 g" = "30 g"`

`m_B = (3 xx "20 g") + (2 xx "10 g") = "80 g"`

`m_C = "20 g" + (7 xx "10 g") = "90 g"`

Now, you know that the ratio for the three chemicals in the final mixture must be `1:3:6`. Keep this in mind.

Let's say that `x` represents the mass of chemical `"B"` in the third mixture, the one that contains `"B"` and `"C"` in a `1:9` ratio.

This means that the mass of `"C"` will be `(9 xx x)`, and the total mass of the third mixture will be

`x + (9 xx x) = 10x`

Now, the mass of chemical `"A"` will not change when you add this third mixture, since it only contains `"B"` and `"C"`. This means that the final mass of `"A"` will be

`m_A = "30 g"`

Now look at the `1:3:6` ratio that must exist between `"A"`, `"B"`, and `"C"`. According to this ratio, the final mixture will contain

`3 xx "30 g" = "90 g" ->` the final mass of chemical `"B"`

`6 xx "30 g" = "180 g" ->` the final mass of chemical `"C"`

This means that you can say, using the mass of `"B"` in the third mixture, `x`, and the mass of `"B"` in the first two mixtures, that

`x + "80 g" = "90 g" implies x = "10 g"`

The same can be said for `"C"`

`9x + "90 g" = "180 g" implies x = "10 g"`

Therefore, the third mixture must contain `"10 g"` of `"B"` and `"90 g"` of `"C"`. The total mass of this third mixture must thus be

`m_"third mixture" = "10 g" + "90 g" = color(green)("100 g")`