Question #11ee0

1 Answer
May 20, 2017

Answer:

#"14.2 g"#

Explanation:

You know that

#"Na"_ 2"CO"_ (3(aq)) + "CuCl"_ (2(aq)) -> "CuCO"_ (3(s)) darr + 2"NaCl"_ ((aq))#

The balanced chemical equation tells you that every #1# mole of sodium carbonate that takes part in the reaction consumes #1# mole of copper(II) chloride.

#"1 mole Na"_2"CO"_3 " " -> " " "1 mole CuCl"_2" "# (mole ratio)

In order to find a relationship between the number of grams of each reactant that takes part in the reactions, you must use the molar masses of the two compounds.

You have

#M_ ("M Na"_ 2"CO"_ 3) = "105.99 g mol"^(-1)#

#M_ ("CuCl"_ 2) = "134.45 g mol"^(-1)#

So, if #1# mole of sodium carbonate has a mass of #"105.99 g"# and #1# mole of copper(II) chloride has a mass of #"134.45 g"#, you can say that you have

#"105.99 g Na"_2"CO"_3 " " -> " " "134.45 g CuCl"_2" "# (gram ratio)

This means that in order for the reaction to completely consume #"18.0 g"# of copper(II) chloride, it must also consume

#18.0 color(red)(cancel(color(black)("g CuCl"_2))) * ("105.99 g Na"_2"CO"_3)/(134.45 color(red)(cancel(color(black)("g CuCl"_2)))) = color(darkgreen)(ul(color(black)("14.2 g Na"_2"CO"_3)))#

The answer is rounded to three sig figs, the number of sig figs you have for the mass of copper(II) chloride.