Answer:
12.6 g
Explanation:
The formula for the calculation of moles is shown below:
![moles = \frac{Mass\ taken}{Molar\ mass}](https://tex.z-dn.net/?f=moles%20%3D%20%5Cfrac%7BMass%5C%20taken%7D%7BMolar%5C%20mass%7D)
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:
![moles = \frac{Mass\ taken}{Molar\ mass}](https://tex.z-dn.net/?f=moles%20%3D%20%5Cfrac%7BMass%5C%20taken%7D%7BMolar%5C%20mass%7D)
Thus,
![Moles= \frac{8.14\ g}{58.12\ g/mol}](https://tex.z-dn.net/?f=Moles%3D%20%5Cfrac%7B8.14%5C%20g%7D%7B58.12%5C%20g%2Fmol%7D)
![Moles\ of\ butane= 0.14\ mol](https://tex.z-dn.net/?f=Moles%5C%20of%5C%20butane%3D%200.14%5C%20mol)
Given: For ![O_2](https://tex.z-dn.net/?f=O_2)
Given mass = 41 g
Molar mass of
= 31.9988 g/mol
The formula for the calculation of moles is shown below:
![moles = \frac{Mass\ taken}{Molar\ mass}](https://tex.z-dn.net/?f=moles%20%3D%20%5Cfrac%7BMass%5C%20taken%7D%7BMolar%5C%20mass%7D)
Thus,
![Moles= \frac{41\ g}{31.9988\ g/mol}](https://tex.z-dn.net/?f=Moles%3D%20%5Cfrac%7B41%5C%20g%7D%7B31.9988%5C%20g%2Fmol%7D)
![Moles\ of\ O_2 = 1.28\ mol](https://tex.z-dn.net/?f=Moles%5C%20of%5C%20O_2%20%3D%201.28%5C%20mol)
According to the given reaction:
![2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O](https://tex.z-dn.net/?f=2C_4H_%7B10%7D%2B13O_2%5Crightarrow%208CO_2%2B10H_2O)
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
= 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