find the amount of c2h6 with 22.4 grams of o2
Answers
Explanation:
We are given:
The mass of ethane gas = 60.0 g
Let us first write the given balanced chemical reaction for the complete combustion of ethane gas.
2
C
2
H
6
+
7
O
2
→
4
C
O
2
+
6
H
2
O
2C2H6+7O2→4CO2+6H2O
From the above-balanced chemical equation, we can see that 2 moles of ethane gas require 7 moles of oxygen gas to completely react with to form 4 moles of carbon dioxide gas and 6 moles of water.
We also know that:
The molar mass of oxygen gas = 32.00 g/mol
The molar mass of ethane gas = 30.07 g/mol
So, we can use the molar masses and the stoichiometry of the reaction to calculate the grams of oxygen needed to combust 60.0 g of ethane:
(
60.0
g
C
2
H
6
)
(
1
m
o
l
C
2
H
6
30.07
g
C
2
H
6
)
(
7
m
o
l
O
2
2
m
o
l
C
2
H
6
)
(
32.00
g
O
2
1
m
o
l
O
2
)
=
223
g
O
2
(60.0gC2H6)(1molC2H630.07gC2H6)(7molO22molC2H6)(32.00gO21molO2)=223gO2
Now, according to the volume-volume relation in a chemical reaction having all its components in the gaseous phase, the volume-volume relation is similar to the mole-mole which is in proportion with the stoichiometric coefficients of the balanced chemical reaction since in the ideal gas law when pressure and temperature are constant, volume of a gas is directly proportional to the number of moles of gas present. We also know that at standard temperature and pressure, one mole of gas will occupy 22.4 L of volume. We will use this to convert volume of gas to moles, apply our stoichiometric mole ratio, and then convert back to volume:
(
20.0
L
C
2
H
6
)
(
1
m
o
l
C
2
H
6
22.4
L
C
2
H
6
)
(
7
m
o
l
O
2
2
m
o
l
C
2
H
6
)
(
22.4
L
O
2
1
m
o
l
O
2
)
=
70.0
L
O
2