Respuesta :
Answer:
12.6 g
Explanation:
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
For butane:-
Mass of butane = 8.14 g
Molar mass of butane = 58.12 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]Moles= \frac{8.14\ g}{58.12\ g/mol}[/tex]
[tex]Moles\ of\ butane= 0.14\ mol[/tex]
Given: For [tex]O_2[/tex]
Given mass = 41 g
Molar mass of [tex]O_2[/tex] = 31.9988 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]Moles= \frac{41\ g}{31.9988\ g/mol}[/tex]
[tex]Moles\ of\ O_2 = 1.28\ mol[/tex]
According to the given reaction:
[tex]2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O[/tex]
2 moles of butane react with 13 moles of oxygen
Also,
1 mole of butane react with 6.5 moles of oxygen
So,
0.14 mole of butane react with 6.5*0.14 moles of oxygen
Moles of oxygen = 0.91 moles
Available moles of [tex]O_2[/tex] = 1.28 moles (Extra)
Limiting reagent is the one which is present in small amount. Thus, butane is limiting reagent.
The formation of the product is governed by the limiting reagent. So,
2 moles of butane forms 10 moles of water
Also,
1 mole of butane forms 10 moles of water
So,
0.14 mole of butane forms 5*0.14 mole of water
Moles of water = 0.7 moles
Molar mass of water = 18 g/mol
So,
Mass of water= Moles × Molar mass = 0.7 × 18 g = 12.6 g
