Lead (II)carbonate decomposes to give lead (II)oxide and carbon dioxide: PbCO₃ (s)→ PbO (s)+ CO₂ (g) If the reaction yield is 95.7%, how many grams of lead (II)oxide will be produced by the decomposition of 2.50 g of lead (II)carbonate? A)1.04 B)1.55 C)2.09 D)4.00 E)5.55

Question

Quizplus · Accepted Answer

First, we need to calculate the theoretical yield of lead (II)oxide:

1 mole of PbCO3 produces 1 mole of PbO

Molar mass of PbCO3 = 267.2 g/mol
1 mole of PbCO3 = 267.2 g

Therefore, 2.50 g of PbCO3 is equal to:

2.50 g PbCO3 x (1 mol PbCO3 / 267.2 g PbCO3) = 0.00936 mol PbCO3

Since the reaction yield is 95.7%, the actual yield of PbO is:

0.0957 x theoretical yield

0.0957 x 0.00936 mol PbO = 0.000895 mol PbO

Finally, we need to convert moles of PbO to grams:

Molar mass of PbO = 223.2 g/mol
0.000895 mol PbO x 223.2 g/mol = 0.1995 g PbO

Therefore, the answer is:

C) 2.09

Lead (II)carbonate Decomposes to Give Lead (II)oxide and Carbon Dioxide