Question

If 15.5 mL of 0.225 M aqueous magnesium chloride is added to 37.5 mL of 0.250 M aqueous lead(II) nitrate, then what mass of lead(II) chloride precipitates

Answer 1

Answer:

0.973 g of PbCl2

Explanation:

MgCl2(aq) + Pb(NO3)2(aq) -----> PbCl2(s) + Mg(NO3)2(aq)

From the information provided

amount of MgCl2 reacted;

15.5/1000 * 0.225 = 0.0035 moles

If 1 mole of MgCl2 yields 1 mole of PbCl2

0.0035 moles of MgCl2 yields 0.0035 moles of PbCl2

For lead nitrate

Amount reacted = 37.5/1000 * 0.250 = 0.0094 moles

If 1 mole of lead nitrate yields 1 mole of lead chloride

0.0094 moles of lead nitrate yields 0.0094 moles of lead chloride

Since MgCl2 is the limiting reactant (it yields the least amount of product)

Then mass of lead chloride produced = 0.0035 moles of PbCl2 * 278.1 g/mol = 0.973 g of PbCl2