2
$\begingroup$

A mixture weighing $4.08\ \mathrm g$ of $\ce{BaO}$ ($M(\ce{Ba})$ = $138\ \mathrm{g\ mol^{-1}}$) and an unknown carbonate $\ce{XCO3}$ was heated strongly. The residue weighed $3.64\ \mathrm g$. This was dissolved in $100\ \mathrm{ml}$ of $1\ \mathrm M$ $\ce{HCl}$. The excess acid required $16\ \mathrm{ml}$ of $2.5\ \mathrm M$ $\ce{NaOH}$ solution for complete neutralization. What is the molar mass of $\ce{X}?$

My approach:

$16\ \mathrm{ml}$ of $2.5\ \mathrm M$ $\ce{NaOH}$ means $0.04\ \mathrm{mol}$ of $\ce{NaOH}$. Thus, the excess acid was $0.04\ \mathrm{mol}$ of $\ce{HCl}$. $100\ \mathrm{ml}$ of $1\ \mathrm M$ $\ce{HCl}$ means $0.1\ \mathrm{mol}$ of $\ce{HCl}$. Thus, $0.06\ \mathrm{mol}$ of $\ce{HCl}$ reacted with the residue. Now, how can I find the ratio in which the residue reacted with $\ce{HCl}?$ What is the composition of the residue?

$\endgroup$
0

1 Answer 1

4
$\begingroup$

In the comments I mentioned one way to solve the problem, but it turns out that was the roundabout way! There is a simpler solution, as follows.

From the neutralization reaction, it is possible to calculate that the mixture reacted with $0.06\ \mathrm{mol}$ of $\ce{H+}$. The carbonate and the oxide react in a proportion of 1:2 with the hydrogen ions, so we know that $n_{\ce{BaO}}+n_{\ce{XO}}=0.03\ \mathrm{mol}$. The mass balance from the thermal decomposition then suggests that $m_{\ce{XCO3}}-m_{\ce{XO}}=0.44\ \mathrm g$. These two equations are already enough.

Notice that the for the molar masses of the unknown oxide and carbonate, we have $M_{\ce{XCO3}}-M_{\ce{XO}}=44\ \mathrm{g/mol}$. If one converts the masses in the second equation into expressions in $M$ and $n$, the result is $n_{\ce{XCO3}}M_{\ce{XCO3}}-n_{\ce{XO}}M_{\ce{XO}}=0.44\ \mathrm g$, which can be turned into $n_{\ce{XCO3}}M_{\ce{XCO3}}-n_{\ce{XO}}\left(M_{\ce{XCO3}}-44\ \mathrm{g/mol}\right)=0.44\ \mathrm g$. It is easy to see that the thermal decomposition does not alter the number of moles of compound containing $\ce{X}$, so that $n_{\ce{XCO3}}=n_{\ce{XO}}$. Thus:

$$n_{\ce{XCO3}}M_{\ce{XCO3}}-n_{\ce{XCO3}}\left(M_{\ce{XCO3}}-44\ \mathrm{g/mol}\right)=0.44\ \mathrm g$$

$$n_{\ce{XCO3}}\left(M_{\ce{XCO3}}-M_{\ce{XCO3}}+44\ \mathrm{g/mol}\right)=0.44\ \mathrm g$$

$$n_{\ce{XCO3}}\times 44\ \mathrm{g/mol}=0.44\ \mathrm g\quad\Rightarrow\quad n_{\ce{XCO3}}=0.01\ \mathrm{mol}$$

Now the problem unravels easily. We get that $n_{\ce{BaO}}=0.02\ \mathrm{mol}$, which implies $m_{\ce{BaO}}=3.08\ \mathrm g$ and $m_{\ce{XCO3}}=1.00\ \mathrm g$. Hence, $M_{\ce{XCO3}}=100\ \mathrm{g/mol}$, and from there $M_{\ce{X}}=40\ \mathrm{g/mol}$. A quick look on a periodic table yields calcium as the most likely candidate, and indeed its carbonate has the formula $\ce{CaCO3}$ and a molar mass of $100\ \mathrm{g/mol}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.