_C4H10O + _O2 =_CO2 + _H2O
Answers
Answer:
We have the unbalanced equation:
C
4
H
10
(
g
)
+
O
2
(
g
)
→
C
O
2
(
g
)
+
H
2
O
(
g
)
This is the combustion of butane,
C
4
H
10
.
Let's first balance the carbons. There are four on the
LHS
, but only one on the
RHS
, so we multiply
C
O
2
by
4
to get:
C
4
H
10
(
g
)
+
O
2
(
g
)
→
4
C
O
2
(
g
)
+
H
2
O
(
g
)
Now, let's balance the hydrogens. There are
10
on the left side, but only two on the right side, so we multiply
H
2
O
by
5
, and get:
C
4
H
10
(
g
)
+
O
2
(
g
)
→
4
C
O
2
(
g
)
+
5
H
2
O
(
g
)
Final step is to balance the oxygens. There are two on the left hand side, but thirteen on the right hand side, so we need to divide thirteen by two to get the "scale number", which is
6.5
. The equation is thus:
C
4
H
10
(
g
)
+
6.5
O
2
(
g
)
→
4
C
O
2
(
g
)
+
5
H
2
O
(
g
)
But wait, we cannot have half a molecule! So, we need to multiply the whole equation by
2
, which leads us to the finalized, balanced equation
Explanation:
Ans is in photo.
C4H100+ 29O2 = 4CO2+ 50H2O
