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%

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

Answer Link

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%

Respuesta :

Otras preguntas

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%