0
$\begingroup$

A 1.61-g sample of a mixture of $\ce{AgNO3}$ and $\ce{NaNO3}$ is treated with excess $\ce{Na2S}\:(\mathrm{aq})$. The precipitate is filtered off, dried and weighed. The dried precipitate weighs 0.44 g. What is the percentage by mass of $\ce{NaNO3}$ in the original mixture?

I attempted this problem by using two equations: one with $\ce{AgNO3}$ and $\ce{Na2S}$, and the other with the other reactant and $\ce{Na2S}$. Then I wrote an equation in moles, using m and 1.61-m as the mass of the reactants. My answer was 2.34 g, which is greater than the original so I know I did something wrong.

$\endgroup$
1
$\begingroup$

The dried precipitate is $\ce{Ag2S}$. The number of moles of $\ce{Ag2S}$ is: $$n=\frac{m_{\ce{Ag2S}}}{M_{\ce{Ag2S}}}=\frac{0.44}{32+108\times 2}=\frac{0.44}{248}=0.00177\,\mathrm{mol}$$ The number of moles of ion $\ce{Ag(I)}$ is$$n'=2n=0.00355\,\mathrm{mol}$$

On the other hand, the percentage by mass of $\ce{AgNO3}$ in the original mixture is: $$\frac{m_{\ce{AgNO3}}}{m_{\ce{(AgNO3 + NaNO3)}}}\times 100=\frac{n'\times M_{\ce{AgNO3}}}{m_{\ce{(AgNO3 +NaNO3)}}}\times 100=\frac{0.00355 \times (108+14+16 \times 3)}{1.61}\times 100=37.48\%$$

So, the percentage by mass of $\ce{NaNO3}$ is: $100-37.48=62.52\%$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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