10. Lead nitrate can be decomposed by heating. What is the percent yield of the decomposition reaction if 15.6 g Pb (NO3) 2 are heated to give 9.4 g of PbO? 2Pb (NO3) 2 (s) → PbO (s) + 4NO2 (g) + O2 (g) *

a. 59%

b. 73%

c. 82%

d. 90%

Question

Chemistry

Bronson Peck

3 August, 12:12

10. Lead nitrate can be decomposed by heating. What is the percent yield of the decomposition reaction if 15.6 g Pb (NO3) 2 are heated to give 9.4 g of PbO? 2Pb (NO3) 2 (s) → PbO (s) + 4NO2 (g) + O2 (g) *

a. 59%

b. 73%

c. 82%

d. 90%

+4

Answers (1)

Know the Answer?

Jorden Hines · Answer 1

First, the equation is unbalanced:

2Pb (NO₃) ₂ → 2PbO + 4NO₂ + O₂

Moles of Pb (NO₃) ₂ = 15.6 / (207 + 2 (14 + 16 x 3))

Moles of Pb (NO₃) ₂ = 0.047

Theoretical yield = 0.047 moles

Theoretical yield = 0.047 x (207 + 16)

Theoretical yield = 10.48

Percentage yield = (9.4. / 10.48) x 100

Percentage yield = 89.6%

The answer is D

10. Lead nitrate can be decomposed by heating. What is the percent yield of the decomposition reaction if 15.6 g Pb (NO3) 2 are heated to give 9.4 g of PbO? 2Pb (NO3) 2 (s) → PbO (s) + 4NO2 (g) + O2 (g) *a. 59%b. 73%c. 82%d. 90%

10. Lead nitrate can be decomposed by heating. What is the percent yield of the decomposition reaction if 15.6 g Pb (NO3) 2 are heated to give 9.4 g of PbO? 2Pb (NO3) 2 (s) → PbO (s) + 4NO2 (g) + O2 (g) *

a. 59%

b. 73%

c. 82%

d. 90%