Question #11ee0
1 Answer
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 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
#"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 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.