Question

A chemist starts with 100 g of Lead (II) sulfate and needs to produce 43 g of Tin (IV) sulfate. How much Tin (IV) dichromate must he use in order to accomplish this goal?

Answer 1

First we need to write out the chemical reaction we're dealing with. In this case it's the following double displacement reaction:

\[2PbSO_{4} + Sn(Cr_{2}O_{7})_{2} \rightarrow 2PbCr_{2}O_{7}+Sn(SO_{4})_{2}\]
The question is really asking you about stoichiometry, so we need numbers. To do this, we need to convert the mass to moles. Since we only need one of the given masses to make our stoichiometric ratio, I'll use the lead (II) sulfate mass.

The molar mass of lead (II) sulfate is 303.26 g/mol. Therefore, we have 100 g / 303.26 g/mol = 0.33 mol of lead (II) sulfate. Now for the ratio to solve for the unknown amount of tin (IV) dichromate (I'll call it x):

\[\frac{ x }{ 0.33 }=\frac{1}{2} \rightarrow x=0.165\]
So we have 0.165 mol of tin (IV) dichromate. We just need to convert this back to a mass now by multiplying by the molar mass of tin (IV) dichromate (550.686 g/mol): 0.165 mol x 550.686 g/mol = 90.8 g.

If you have any questions about any of this let me know!