# Calculate the mass of the chalk that contains impurities?

Aug 1, 2017

The equation is.....

$C a C {O}_{3} \left(s\right) + 2 H C l \rightarrow C a C {l}_{2} \left(a q\right) + {H}_{2} O \left(l\right) + C {O}_{2} \left(g\right) \uparrow$

I get a purity of 90%.......

#### Explanation:

We need to work out $\left(i\right)$ the STARTING number of moles of acid......

$= 0.100 \cdot L \times 0.50 \cdot m o l \cdot {L}^{-} 1 = 0.050 \cdot m o l \cdot H C l$

Now this quantity $\left(i i\right)$ reacted with the calcium carbonate in the chalk, and $\left(i i i\right)$ the EXCESS quantity was treated with $28.0 \cdot m L$ of $0.50 \cdot m o l \cdot {L}^{-} 1$ $N a O H$, i.e. $28.0 \times {10}^{-} 3 \cdot L \times 0.50 \cdot m o l \cdot {L}^{-} 1$
$= 0.014 \cdot m o l$.

We need $\left(i i\right)$, which given the above, $\left(i i\right) = \left(i\right) - \left(i i i\right)$

$\equiv 0.050 \cdot m o l - 0.014 \cdot m o l = 0.036 \cdot m o l$

And thus given the 2:1 stoichiometry of the carbonate reaction, there were $\frac{0.036 \cdot m o l}{2} = 0.018 \cdot m o l$ with respect to $C a C {O}_{3} \left(s\right)$.......

i.e. a mass of $0.018 \cdot m o l \times 100.09 \cdot g \cdot m o {l}^{-} 1 = 1.80 \cdot g$

And thus "% purity"="Mass of calcium carbonate"/"Mass of chalk"xx100%

(1.80*g)/(2.00*g)xx100%=90%; and so the calcium carbonate was 10% by mass impure.

We have assume that the impurities were NOT basic, which would skew the results. Please go over my calculations, and point out any errors or misapprehensions (all care taken but no responsibility admitted!). If there is a step which I have glossed over or ignored, please point it out.