Answer:
Reactant : A combustion of hydrocarbon.
Explanation:
It is known that when hydrocarbons are involved in combustion, they produce carbon dioxide and water.
CxHy + (x+y/4)O2 ===> xCO2 + y/2H2O
10 x 70 = 100 x Part, or
700 = 100 x Part
Now, divide by 100 and get the answer:
Part = 700 / 100 = <span>7</span>
On complete conversion (100% yield) 9.75 g of ethyl butyrate will be produced. Below is the solution.....
Answer : The molecule of ethane present in 64.28 L of ethane gas at STP is,
molecule.
Solution :
At STP,
22.4 L volume of ethane present in 1 mole of ethane gas
64.28 L volume of ethane present in
of ethane gas
And, as we know that
1 mole of ethane molecule contains
molecules of ethane
2.869 moles of ethane molecule contains
molecules of ethane
Therefore, the molecule of ethane present in 64.28 L of ethane gas at STP is,
molecule.