Lead nitrate can be decomposed by heating. What is the percent yield of the decomposition reaction if 9.9 g Pb(NO3)2 are heated to give 5.5 g of PbO? ROUND TO THE NEAREST TENTH!!
2Pb(NO3)2(s) → 2PbO(s) + 4NO2(g) + O2(g)
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Explanation:
We need (i) a stoichiometric equation........
P
b
(
N
O
3
)
2
+
Δ
→
P
b
O
(
s
)
+
2
N
O
2
(
g
)
+
1
2
O
2
(
g
)
This simply must be known; nitrates undergo thermal decomposition to give oxide and nitric oxide and dioxygen gas.
And (ii). given the 1:1 stoichiometry between nitrate and oxide, we needs equivalent quantities of lead oxide and lead nitrate:
Percentage yield
=
Moles of lead oxide
Moles of lead nitrate
×
100
%
=
5.5
⋅
g
223.2
⋅
g
⋅
m
o
l
−
1
9.9
⋅
g
331.2
⋅
g
⋅
m
o
l
−
1
×
100
%
=
?
?
%
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