Question

Lead nitrate can be decomposed by heating. What is the percent yield of the decomposition reaction if 9.9 g of Pb(NO3)2 are heated to give 5.5 g of PbO? 2Pb(NO3)2(s) → 2PbO(s) + 4NO2(
g. + O2(
g.

Answer 1

Answer:

Percentage yield is 82.44%

Explanation:

Percentage yield is the ratio of actual product produced to theoretical product produced.

Step 1: Determine the actual yield of the reaction

The number of moles of $PbO$ are determined using the below formula where 'n' is the number of moles, 'm' is the mass and 'M' is the molar mass

$n=\frac{m}{M}$

$n_{ay,PbO}=\frac{m_{PbO}}{M_{PbO}}$

$n_{ay,PbO}=\frac{5.5}{207.2+16}$

$n_{ay,PbO}=\frac{5.5}{223.2}$

$n_{ay,PbO}=0.02464$

The number of moles of $PbO$ give the actual yield of the reaction

Step 2: Determine the theoretical yield of the reaction

The same formula for number of moles is used to determine the moles of $Pb(NO_{3})_{2}$

$n_{Pb(NO_{3})_{2}}=\frac{m_{Pb(NO_{3})_{2}}}{M_{Pb(NO_{3})_{2}}}$

$n_{Pb(NO_{3})_{2}}=\frac{9.9}{207.2+(14+16*3)*2}$

$n_{Pb(NO_{3})_{2}}=\frac{9.9}{331.2}$

$n_{Pb(NO_{3})_{2}}=0.02989$

Based on the balanced chemical equation given in the question, one mole of $PbO$ should be produced for every mole of $Pb(NO_{3})_{2}$ decomposed. Thus,

$n_{ty,PbO}=0.02989$

Step 3: Determine percentage yield

$%yield = \frac{actual yield}{theoretical yield}*100$

$%yield = \frac{n_{ay,PbO}}{n_{ty,PbO}}*100$

$%yield = \frac{0.02464}{0.02989}*100$

$%yield = 82.44%$

Answer 2

2Pb(NO₃)₂ = 2PbO + 4NO₂ + O₂

m{Pb(NO₃)₂}=9.9 g
m'{PbO}=5.5 g
M{Pb(NO₃)₂}=331.2 g/mol
M{PbO}=223.2 g/mol
w-?

w=m'{PbO}/m{PbO}

m{PbO}=M{PbO}m{Pb(NO₃)₂}/M{Pb(NO₃)₂}

w=m'{PbO}M{Pb(NO₃)₂}/[M{PbO}m{Pb(NO₃)₂}]

w=5.5*331.2/[223.2*9.9]=0.8244 (82.44%)