Answer:
B. 0.971 g
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:
<em>Moles MgCl₂:</em>
15.5mL = 0.0155L * (0.225 mol / L) = 3.49x10⁻³ moles
<em>Moles Pb(NO₃)₂:</em>
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:
<h3>B. 0.971 g</h3>
