Answer:
To solve this problem, we need to first determine the limiting reactant, which is the reactant that will be completely consumed in the reaction, thereby limiting the amount of product that can be formed.
We'll use the given masses of butane and oxygen to find out which one is the limiting reactant.
Calculate the molar masses of butane (
�
4
�
10
C
4
H
10
) and oxygen (
�
2
O
2
):
Molar mass of butane (
�
4
�
10
C
4
H
10
) = 4(12.01 g/mol) + 10(1.008 g/mol) ≈ 58.12 g/mol
Molar mass of oxygen (
�
2
O
2
) = 2(16.00 g/mol) ≈ 32.00 g/mol
Calculate the number of moles of each reactant:
Moles of butane = Mass of butane / Molar mass of butane = 42 g / 58.12 g/mol ≈ 0.722 mol
Moles of oxygen = Mass of oxygen / Molar mass of oxygen = 71.4 g / 32.00 g/mol ≈ 2.23 mol
Now, we need to determine which reactant is limiting by comparing the moles of each reactant to the stoichiometric coefficients in the balanced chemical equation for the reaction:
�
4
�
10
+
13
/
2
�
2
→
4
�
�
2
+
5
�
2
�
C
4
H
10
+13/2O
2
→4CO
2
+5H
2
O
From the balanced equation, we see that 1 mole of butane (
�
4
�
10
C
4
H
10
) reacts with 13/2 moles of oxygen (
�
2
O
2
).
We calculate the moles of oxygen required for the given moles of butane:
Moles of oxygen required
=
Moles of butane
×
13
2
Moles of oxygen required=Moles of butane×
2
13
=
0.722
×
13
2
=0.722×
2
13
≈
4.703
mol
≈4.703 mol
Since the moles of oxygen available (2.23 mol) are less than the moles required (4.703 mol), oxygen is the limiting reactant.
Now, we calculate the mass of water produced using the limiting reactant:
From the balanced equation, we see that 1 mole of oxygen produces 5 moles of water.
Moles of water produced
=
Moles of oxygen
×
5
Moles of water produced=Moles of oxygen×5
=
2.23
×
5
=2.23×5
=
11.15
mol
=11.15 mol
Now, we calculate the mass of water produced:
Mass of water
=
Moles of water
×
Molar mass of water
Mass of water=Moles of water×Molar mass of water
=
11.15
mol
×
(
2
×
1.008
�
/
�
�
�
+
16.00
�
/
�
�
�
)
=11.15 mol×(2×1.008g/mol+16.00g/mol)
≈
180.0
g
≈180.0 g
So, the maximum mass of water that could be produced by the chemical reaction is approximately 180.0 grams.
Explanation:
