# Question 65722

Feb 20, 2016

$\text{100 g}$

#### Explanation:

You are on the right track with this one.

Use the known ratio for the first mixture of $\text{A}$, $\text{B}$, and $\text{C}$ to determine the mass of each chemical in the $\text{100-g}$ sample.

If you take ${m}_{1}$ to be the mass of chemical $\text{A}$, you can say that

$3 \times {m}_{1} \to$ the mass of chemical $\text{B}$

$1 \times {m}_{1} \to$ the mass of chemical $\text{C}$

This means that you have

${m}_{1} + 3 \cdot {m}_{1} + {m}_{1} = \text{100 g}$

$5 \cdot {m}_{1} = \text{100 g" implies m_1 = "100 g"/5 = "20 g}$

Do the same for the second $\text{100-g}$ mixture of $\text{A}$, $\text{B}$, and $\text{C}$. Let's say that ${m}_{2}$ is the mass of chemical $\text{A}$ in this mixture. You will have

$2 \times {m}_{2} \to$ the mass of chemical $\text{B}$

$7 \times {m}_{2} \to$ the mass of chemical $\text{C}$

This means that you have

${m}_{2} + 2 \cdot {m}_{2} + 7 \cdot {m}_{2} = \text{100 g}$

$10 \cdot {m}_{2} = \text{100 g" implies m_2 = "10 g}$

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

${m}_{A} = \text{20 g" + "10 g" = "30 g}$

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

${m}_{C} = \text{20 g" + (7 xx "10 g") = "90 g}$

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 $\text{B}$ in the third mixture, the one that contains $\text{B}$ and $\text{C}$ in a $1 : 9$ ratio.

This means that the mass of $\text{C}$ will be $\left(9 \times x\right)$, and the total mass of the third mixture will be

$x + \left(9 \times x\right) = 10 x$

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

${m}_{A} = \text{30 g}$

Now look at the $1 : 3 : 6$ ratio that must exist between $\text{A}$, $\text{B}$, and $\text{C}$. According to this ratio, the final mixture will contain

$3 \times \text{30 g" = "90 g} \to$ the final mass of chemical $\text{B}$

$6 \times \text{30 g" = "180 g} \to$ the final mass of chemical $\text{C}$

This means that you can say, using the mass of $\text{B}$ in the third mixture, $x$, and the mass of $\text{B}$ in the first two mixtures, that

$x + \text{80 g" = "90 g" implies x = "10 g}$

The same can be said for $\text{C}$

$9 x + \text{90 g" = "180 g" implies x = "10 g}$

Therefore, the third mixture must contain $\text{10 g}$ of $\text{B}$ and $\text{90 g}$ of $\text{C}$. The total mass of this third mixture must thus be

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