Question #65722

1 Answer
Feb 20, 2016

Answer:

#"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")#