If you have 381 g Cu and 1359 g AgNO3, how many grams of silver could be made?
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864g
C
u
(
s
)
+
2
A
g
N
O
3
(
a
q
)
→
C
u
(
N
O
3
)
2
(
a
q
)
+
2
A
g
⏐
⏐
↓
1
mol
+
2
mol
→
2
mol
Now we have 281g Cu = 381/63.5 = 6 mol.
We have 1359g silver nitrate = 1359/170 = 8 mol
But 6 mol Cu requires 12 mol silver nitrate.
So it is the 8 mol of silver nitrate which will decide the yield of the reaction since Cu is in XS.
So 8 mol silver nitrate
→
8 mol Ag
8 mol Ag = 108 x 8 = 864g
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