Question

When H2SO4 is added to PbI2, a precipitate of PbSO4 forms. The PbSO4 is then filtered from the solution, dried, and weighed. If the recovered PbSO4 is found to have a mass of 0.0471 g, how many moles of iodide ions were in the original solution?

Answer 1

Answer:

$n_{I^-}=3.11x10^{-4}molI^-$

Explanation:

Hello,

In this case, the undergoing chemical reaction is shown below:

$H_2SO_4(aq)+PbI_2(aq)\rightarrow PbSO_4(s)+2HI(aq)$

Thus, we obtain 0.0471 g of lead (II) sulfate, the iodide ions  in the original solution are computed below by stoichiometry, taking into account that into 1 mole of lead (II) iodide there are 2 moles of iodide ions:

$n_{I^-}=0.0471gPbSO_4*\frac{1molPbSO_4}{303.26gPbSO_4}*\frac{1molPbI_2}{1molPbSO_4}*\frac{2molI^-}{1molPbI_2} \\n_{I^-}=3.11x10^{-4}molI^-$

Best regards.