How many grams of oxygen are required to burn 13.50 grams of acetylene?

About
553.6
grams of oxygen are used in this combustion.
Explanation:
Molecular mass of acetylene:
2
⋅
M
M
C
+
2
⋅
M
M
H

M
M
C
= Molecular mass of carbon
M
M
H
= Molecular mass of hydrogen
So,
2
⋅
12
+
2
⋅
1
=
26
grams/mole
Mass used in combustion
molecular mass
=
moles used in combustion
180
grams
26
grams
mole
=
6.92
moles
According to the main equation, to 2 moles of acetylene, 5 moles of oxygen are used.
M
M
O
=
2
⋅
16
=
32
grams/mole
So, if we use
6.92
moles of Acetylene,
x
moles of oxygen will be necessary.
x
=
6.92
⋅
5
2
=
3.46
⋅
5
=
17.3
moles of oxygen.
17.3
moles
⋅
32
grams
mole
=
553.6
grams of oxygen

The answer is c. 320 g