Answer :
Answer: The maximum amount of water that could be produced by the chemical reaction is 20.16 grams
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
- For butane:
Given mass of butane = 13 g
Molar mass of butane = 58.12 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of butane}=\frac{13g}{58.12g/mol}=0.224mol[/tex]
- For oxygen gas:
Given mass of oxygen gas = 70.9 g
Molar mass of oxygen gas = 32 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of oxygen gas}=\frac{70.9g}{32g/mol}=2.216mol[/tex]
The chemical equation for the reaction of butane and oxygen gas follows:
[tex]2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O[/tex]
By Stoichiometry of the reaction:
2 moles of butane reacts with 13 moles of oxygen gas
So, 0.224 moles of butane will react with = [tex]\frac{13}{2}\times 0.224=1.456mol[/tex] of oxygen gas
As, given amount of oxygen gas is more than the required amount. So, it is considered as an excess reagent.
Thus, butane is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
2 moles of butane produces 10 moles of water
So, 0.224 moles of butane will produce = [tex]\frac{10}{2}\times 0.224=1.12moles[/tex] of water
Now, calculating the mass of water from equation 1, we get:
Molar mass of water = 18 g/mol
Moles of water = 1.12 moles
Putting values in equation 1, we get:
[tex]1.12mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(1.12mol\times 18g/mol)=20.16g[/tex]
Hence, the maximum amount of water that could be produced by the chemical reaction is 20.16 grams