How many grams of NaOH is required to neutralize 54.75 grams of HCL?many
Answers
Answer:
Explanation:
A Neutralization reaction occurs when a strong acid reacts with a strong base to produce water and a salt (ionic compound).
In our case, sulfuric acid (strong acid) will react with sodium hydroxide (strong base) to form water and sodium sulfate:
H
2
S
O
4
+
2
N
a
O
H
→
N
a
2
S
O
4
+
2
H
2
O
We start with the units that we want to end up with and we set that equal to our given value, which will be multiplied by a conversion factor (the molar ratio from the balanced chemical equation). The
98.08
g
m
o
l
represents the molar mass of sulfuric acid and the
40.0
g
m
o
l
is the molar mass of sodium hydroxide.
So, what we have is:
g
N
a
O
H
=
25.0
g
H
2
S
O
4
×
1
m
o
l
H
2
S
O
4
98.08
g
H
2
S
O
4
×
2
m
o
l
N
a
O
H
1
m
o
l
H
2
S
O
4
×
40.0
g
N
a
O
H
1
m
o
l
N
a
O
H
=
Answer:
For neutralization,
moles of NaOH = moles of HCL
w/40 = 54.75/36.5
w = 60g