How many grams of PbCl₂ are formed when 25.0 mL of 0.654 M KCl react with Pb(NO₃)₂ ?
2KCl(aq) + Pb(NO₃) 2(aq) → 2KNO₃(aq) + PbCl₂(s)
A)22.7 g
B)1.64 g
C)2.27 g
D)4.54 g
E)9.08 g

Question

How many grams of PbCl₂ are formed when 25.0 mL of 0.654 M KCl react with Pb(NO₃)₂ ? 
2KCl(aq) + Pb(NO₃) 2(aq) → 2KNO₃(aq) + PbCl₂(s)
A)22.7 g
B)1.64 g
C)2.27 g
D)4.54 g
E)9.08 g

Quizplus · Accepted Answer

1. Calculate the number of moles of KCl:
n = c × V = 0.654 mol/L × 0.0250 L = 0.0164 mol

2. According to the balanced chemical equation, 2 moles of KCl react with 1 mole of Pb(NO₃)₂ to produce 1 mole of PbCl₂.
So, the number of moles of PbCl₂ produced in the reaction is 0.5 × 0.0164 = 0.0082 mol.

3. Finally, use the molar mass of PbCl₂ to convert the number of moles to grams:
m = n × M = 0.0082 mol × 278.1 g/mol = 2.27 g

Therefore, the answer is C) 2.27 g.

How Many Grams of PbCl₂ Are Formed When 25