# Question 11ee0

May 20, 2017

$\text{14.2 g}$

#### Explanation:

You know that

${\text{Na"_ 2"CO"_ (3(aq)) + "CuCl"_ (2(aq)) -> "CuCO"_ (3(s)) darr + 2"NaCl}}_{\left(a q\right)}$

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.

$\text{1 mole Na"_2"CO"_3 " " -> " " "1 mole CuCl"_2" }$ (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 $\text{105.99 g}$ and $1$ mole of copper(II) chloride has a mass of $\text{134.45 g}$, you can say that you have

$\text{105.99 g Na"_2"CO"_3 " " -> " " "134.45 g CuCl"_2" }$ (gram ratio)

This means that in order for the reaction to completely consume $\text{18.0 g}$ of copper(II) chloride, it must also consume

$18.0 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{g CuCl"_2))) * ("105.99 g Na"_2"CO"_3)/(134.45 color(red)(cancel(color(black)("g CuCl"_2)))) = color(darkgreen)(ul(color(black)("14.2 g Na"_2"CO}}_{3}}}}$

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