Answer:
3.13g
Explanation:
First, we'll begin by writing a balanced equation between the reaction of gaseous ethane with gaseous oxygen to produce gaseous carbon dioxide and gaseous water. This is illustrated below:
2C2H6 + 7O2 —> 4CO2 + 6H2O
Next, let us calculate the mass of C2H6 and the mass of O2 that reacted from the balanced equation. This is illustrated below:
Molar Mass of C2H6 = (12x2) + (6x1) = 24 + 6 = 30g/mol
The mass of C2H6 that reacted from the balanced equation = 2 x 30 = 60g
Molar Mass of O2 = 16x2 = 32g/mol
Mass of O2 that reacted from the balanced equation = 7 x 32 = 224g
Now, to calculate the left over mass of ethane (C2H6), let us first calculate the mass of ethane (C2H6) that will react with 22g of oxygen(O2). This can be achieved by doing the following:
From the balanced equation above,
60g of C2H6 reacted with 224g of O2.
Therefore, Xg of C2H6 will react with 22g of O2 i.e
Xg of C2H6 = (60 x 22)/224
Xg of C2H6 = 5.89g
From the calculations above, 5.89g will react completely with 22g of O2.
The left over mass of C2H6 can be obtained as follow:
Mass of C2H6 from the question = 9.02g
Mass of C2H6 that reacted = 5.89g
The left over mass of C2H6 =?
The left over mass of C2H6 = Mass of C2H6 - mass of C2H6 that reacted
Left over mass of C2H6 = 9.02 - 5.89
Left over mass of C2H6 = 3.13g
