How many mililitres of 0.112M of HCl that will react with sodium carbonate in 21.2 mL of 0.150M Na2CO3 according to the following equation?
2HCl + Na2CO3 2NaCl + CO2 + H2O
Answers
Answered by
1
Answer:
3330ml
Explanation:
Explanation:
From the chemical equation, we can see that for every mole of
C
a
C
O
3
, 2 moles of HCl react. This means that if you want them to react exactly in proportion, the number of moles of HCl needed is double that of
C
a
C
O
3
.
n
(
H
C
l
)
=
2
⋅
n
(
C
a
C
O
3
)
The molar mass (MM) of calcium carbonate is:
M
M
(
C
a
C
O
3
)
=
40.1
+
12.0
+
3
⋅
16.0
=
100.1
g
m
o
l
The number of moles of calcium carbonate can now be calculated:
n
(
C
a
C
O
3
)
=
m
a
s
s
M
M
=
51.0
100.1
=
0.509
m
o
l
So, now we double the answer to get the amount of HCl needed:
n
(
H
C
l
)
=
2
⋅
n
(
C
a
C
O
3
)
=
2
⋅
0.509
=
1.02
m
o
l
Rearranging the concentration formula we can get the volume needed in litres:
C
=
n
V
⇒
V
=
n
C
V
=
1.02
0.306
=
3.33
L
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