13 A mass of 2.50g of sodium chloride reacts with excess lead (II) nitrate solution forming lead(II) chloride with a yield of 95%. What is the mass of lead (II) chloride, PbCl2, formed?
2NaCl + Pb (NO3)2 → PbCl2 + 2NaNO3
[Mr values: NaCl = 58.5 PbCl2 = 278.2]
A 5.65g
B 5.94g
C 11.3g
D 11.9g
ANSWER
B 5.94g
Analysis of Option/Explanation
Step 1: Calculating moles of sodium chloride by ration method
1 mole of NaCl contain = 58.5 formula mass unit
X mole of NaCl when= 2.5o g
Therefore,
“X” moles of NaCl = 0.042735 moles
Step 2: Finding moles of PbCl2 in relation to limiting reactant i.e. NaCl
Step 3: Calculating mass of PbCl2
1 mole of PbCl2 contain = 278.2 grams
0.0213 moles when= x gram
Therefore,
X gram of PbCl2 = 5.499 g.
Hence, option B is the correct answer.
