Question

Mole of oxygen required to react completely with 90
g of C2H6 is
2C2H6 + 702 - 4CO2 + 6H20​

Answer 1

Answer:

10.5 moles

Explanation:

number of moles in 90 gram of C2H6 = 90/30 = 3

2 moles of C2H6 reacts with 7 moles of co2 completely,

so, 3 moles of C2H6 reacts with moles of O2 = (7/2) * 3 = 10.5 moles

Answer 2

Given:

The given reaction is $2C_2H_6+7O_2\rightarrow 4CO_2+6H_2O$.

To find:

A mole of oxygen is required to react completely with 90 grams of ethane.

Solution:

$2C_2H_6+7O_2\rightarrow 4CO_2+6H_2O$.

The atomic weight of C is 12 and H is 1 gram.

Hence the molecular weight of ethane is 12×2+6×1=30 grams.

1 mole of ethane is equal to 30 grams of ethane so, 90 grams of ethane is equal to 90/30= 3 moles.

Hence mole of ethane reacts with 7 moles of oxygen.

So, 3 moles of ethane will react with 7×3/2=10.5 moles of oxygen gas.

So, 3 moles of ethane will react with 10.5 moles of oxygen gas.