For the balanced equation, how many grams of SO2 will react with 42 mol of O2?
S + O2 --> SO2
A. 43.1grams
B. 1,274 grams
C. 2,690.5 grams
D. 19.8 grams
Answers
Your answer is ...................
I got
2.5
g of
S
O
3
was produced.
Explanation:
Well, we first write down the balanced equation, that is
2
S
O
2
(
g
)
+
O
2
(
g
)
→
2
S
O
3
(
g
)
From here, we can see that the mole ratio between
S
O
2
and
S
O
3
is
2
:
2
=
1
. That means, for every mole of sulfur dioxide
(
S
O
2
)
used, one mole of sulfur trioxide
(
S
O
3
)
will be produced.
Here, we have used
2
g
of
S
O
2
, so we need to convert that amount into moles.
Sulfur dioxide has a molar mass of
64
g/mol
. So here, we have used
2
g
64
g
/mol
=
0.03125
mol of
S
O
2
I will not yet round off my answer, as we have not gotten the required answer.
Since the mole ratio was
1
, then
0.03125
mol
of
S
O
3
will be produced. We now need to convert that amount into grams, so we can do that by multiplying by the molar mass of sulfur trioxide.
Sulfur trioxide has a molar mass of
80
g/mol
. So here, there exist
0.03125
mol
⋅
80
g/
mol
=
2.5
g of
S
O
3
So,
2.5
grams of sulfur trioxide was produced.