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?A. 0.485 g.B. 0.971 g.C. 1.94 g.D. 3.888 g.E. 5.82 8 g.

Answer 1

B. 0.971 g

Step-by-step explanation:

When MgCl₂(aq) reacts with Pb(NO₃)₂(aq), PbCl₂(s) and Mg(NO₃)₂(aq) are produced:

MgCl₂(aq) + Pb(NO₃)₂(aq) →, PbCl₂(s) + Mg(NO₃)₂(aq)

Thus, we need to find imiting reactant finding moles of each reactant:

Moles MgCl₂:

15.5mL = 0.0155L * (0.225 mol / L) = 3.49x10⁻³ moles

Moles Pb(NO₃)₂:

37.5mL = 0.0375L * (0.250mol / L) = 9.38x10⁻³ moles

As the ratio of the reactants is 1:1, the moles of PbCl₂ are 3.48x10⁻³ moles.

We need to convert thes moles to mass using molar mass of PbCl₂ (278.1g/mol), thus:

3.48x10⁻³ moles * (278.1g/mol) =

0.968g of PbCl₂ are precipitate

Thus, right answer is:

B. 0.971 g