Question d9700

Aug 11, 2016

$\text{0.089 g}$

Explanation:

Start by writing the balanced chemical equations that describe your reactions

${\text{CaCO"_ (3(s)) + color(red)(2)"HCl"_ ((aq)) -> "CaCl"_ (2(aq)) + "H"_ 2"O"_ ((l)) + "CO}}_{2 \left(g\right)} \uparrow$

${\text{MgCO"_ (2(s)) + color(blue)(2)"HCl"_ ((aq)) -> "MgCl"_ (2(aq)) + "H"_ 2"O"_ ((l)) + "CO}}_{2 \left(g\right)} \uparrow$

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 $x$ moles of calcium chloride, ${\text{CaCl}}_{2}$, and the second reaction produced $y$ moles of magnesium chloride, ${\text{MgCl}}_{2}$.

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 $\text{0.19 g}$ of solid product. You can thus say that

$111 \cdot x + 95.2 \cdot y = 0.19 \text{ " " } \textcolor{\mathmr{and} a n \ge}{\left(1\right)}$

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 $\textcolor{red}{2}$ moles of hydrochloric acid and produces $1$ mole of calcium chloride. In your case, the reaction produced $x$ moles of calcium chloride, which means that it must have consumed $\left(\textcolor{red}{2} \cdot x\right)$ moles of hydrochloric acid.

The same can be said about the second reaction. You get $y$ moles of magnesium chloride, which means that the reaction must have consumed $\left(\textcolor{b l u e}{2} \cdot y\right)$ moles of hydrochloric acid.

You can thus say that

$\textcolor{red}{2} \cdot x + \textcolor{b l u e}{2} \cdot y = 0.003696$

which is equivalent to

$x + y = 0.001848 \text{ " " } \textcolor{\mathmr{and} a n \ge}{\left(2\right)}$

All you have to do now is solve for $x$, the number of moles of calcium chloride present in the solid product.

As you can see, calcium carbonate and calcium chloride are in present in a $1 : 1$ mole ratio in the first chemical equation, which means that the number of moles of calcium chloride produced will be equal to the number of moles of calcium carbonate consumed.

Use equation $\textcolor{\mathmr{and} a n \ge}{\left(2\right)}$ to write

$y = 0.001848 - x$

Plug this into equation $\textcolor{\mathmr{and} a n \ge}{\left(1\right)}$ to find

$111 \cdot x + 95.2 \cdot \left(0.001848 - x\right) = 0.19$

$111 \cdot x + 0.176 - 95.2 \cdot x = 0.19$

$15.8 \cdot x = 0.0140 \implies x = \frac{0.0140}{15.8} = 0.0008861$

You can now say that the original mixture contained $0.0008861$ moles of calcium carbonate. Use the compound's molar mass to convert this to grams

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.