Question #d9700
1 Answer
Explanation:
Start by writing the balanced chemical equations that describe your reactions
#"CaCO"_ (3(s)) + color(red)(2)"HCl"_ ((aq)) -> "CaCl"_ (2(aq)) + "H"_ 2"O"_ ((l)) + "CO"_ (2(g)) uarr#
#"MgCO"_ (2(s)) + color(blue)(2)"HCl"_ ((aq)) -> "MgCl"_ (2(aq)) + "H"_ 2"O"_ ((l)) + "CO"_ (2(g)) uarr#
Take a second to look up the molar masses of calcium chloride and magnesium chloride
#M_("M CaCl"_2) = "111 g mol"^(-1)#
#M_("M MgCl"_2) = "95.2 g mol"^(-1)#
Now, let's assume that the first reaction produced
The masses of the two salts can be written using the number of moles and their respective molar masses
#x color(red)(cancel(color(black)("moles CaCl"_2))) * "111 g"/(1color(red)(cancel(color(black)("mole CaCl"_2)))) = (111 * x)color(white)(a) "g CaCl"_2#
#y color(red)(cancel(color(black)("moles MgCl"_2))) * "95.2 g"/(1color(red)(cancel(color(black)("mole MgCl"_2)))) = (95.2 * y)color(white)(a)"g MgCl"_2#
You know that after you evaporate the water from the resulting solution and filtrate the two salts, you end up with a total mass of
#111 * x + 95.2 * y = 0.19" " " "color(orange)((1))#
Next, use the molarity and volume of the hydrochloric acid solution to calculate how many moles were consumed by the two reactions
#42 color(red)(cancel(color(black)("cm"^3))) * (1 color(red)(cancel(color(black)("dm"^3))))/(10^3color(red)(cancel(color(black)("cm"^3)))) * "0.088 moles HCl"/(1color(red)(cancel(color(black)("dm"^3)))) = "0.003696 moles HCl"#
Take the first reaction. Notice that the reaction consumes
The same can be said about the second reaction. You get
You can thus say that
#color(red)(2) * x + color(blue)(2) * y = 0.003696#
which is equivalent to
#x + y = 0.001848" " " "color(orange)((2))#
All you have to do now is solve for
As you can see, calcium carbonate and calcium chloride are in present in a
Use equation
#y = 0.001848 - x#
Plug this into equation
#111 * x + 95.2 * (0.001848 - x) = 0.19#
#111 * x + 0.176 - 95.2 * x = 0.19#
#15.8 * x = 0.0140 implies x= 0.0140/15.8 = 0.0008861#
You can now say that the original mixture contained
#0.0008861 color(red)(cancel(color(black)("moles CaCO"_3))) * "100.1 g"/(1color(red)(cancel(color(black)("mole CaCO"_3)))) = color(green)(|bar(ul(color(white)(a/a)color(black)("0.089 g")color(white)(a/a)|)))#
The answer is rounded to two sig figs.