How many grams of calcium carbonate are needed to form 482 grams of calcium carbide?
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Answer:
We are told that
CaCO
3
is in excess, so clearly,
C
is the limiting reactant. Therefore, since
5 mols C
form
2 mols CaC
2
, we scale the reaction down to get:
4
5
×
(
2
CaCO
3
(
s
)
+
5
C
(
gr
)
Δ
−−→
2
CaC
2
(
s
)
+
3
CO
2
(
g
)
)
Now we read directly from the re-scaled balanced reaction that:
⇒
4
5
×
5 mols C
forms
4
5
×
2
=
8
5
mols CaC
2
.
If the molar mass of
CaC
2
is
64.099 g/mol
, how many grams of
CaC
2
involves
8
5
mols
of it? Your answer must be larger than
64.099 g
to make physical sense.
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