Question

6g of ethane is burnt in excess of O2 the moles of H20fkrmed would be

Answer 1

Hello Dear.

Here is the answer---

The Combustion of the ethane will be represented by the Equation---

2C₂H₆ + 7O₂ -------→ 4CO₂ + 6H₂O

Molar Mass of Ethane = 30 g/mole.
Molar mass of water = 18 g/mole.

From the Equation,

∵ 2 × 30 grams of Ethane produces 6 × 18 grams of Water.
∴ 1 grams of the Ethane produces (9/5) grams of Water.
∴ 6 grams of Ethane produces 10.8 grams of Water.


Now, Using the Formula,
 No. of Moles of Water = Mass/Molar Mass
 = 10.8/18
 = 0.6 moles.


Thus, Number of moles of water produced by 6 grams of ethane is 0.6.


Hope it helps. 

Answer 2

The balanced chemical equation for the above reaction is given by:
2 $C_{2} H_{6}$ + 3 $O_{2}$   --->   4 C + 6  $H_{2} O$ 


The molar mass of Ethane is 30 g/mol and that of Water is 18 g/mol. 

From the above equation, we can observe that
2 * 30 grams of Ethane produces 6 * 18 grams of Water.
Thus,
60 grams of Ethane produces 108 grams of Water.

We can write this in the following way:
60 g Ethane  ---> 108 g Water
Divide both sides by 10
6 g Ethane ---> 10.8 g Water

Thus, on burning 6g of Ethane, we get 10.8 g of water.

No. of moles of Water ( $H_{2} O$ ) formed is given by:
No. of moles = Gram Mass / Molar Mass
No. of moles = 10.8 / 18
No. of moles = 0.6

Thus, 0.6 mole of $H_{2} O$ is formed.