Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 2.71 g of ethane is mixed with 7.6 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to significant digits.

6g

Explanation:

Step 1:

The balanced equation for the reaction between gaseous ethane and gaseous oxygen. This is given below:

2C2H6 + 7O2 —> 4CO2 + 6H2O

Step 2:

Determination of the masses of C2H6 and O2 that reacted and the mass of CO2 produced from the balanced equation. This is illustrated below:

2C2H6 + 7O2 —> 4CO2 + 6H2O

Molar Mass of C2H6 = (12x2) + (6x1) = 24 + 6 = 30g/mol

Mass of C2H6 from the balanced equation = 2 x 30 = 60g

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 7 x 32 = 224g

Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol

Mass of CO2 from the balanced equation = 4 x 44 = 176g

From the balanced equation above,

60g of C2H6 reacted with 224g of O2 to produce 176g of CO2.

Step 3:

Determination of the limiting reactant.

We need to determine the limiting reactant as it will be needed to obtain the maximum mass of CO2.

From the balanced equation above,

60g of C2H6 reacted with 224g of O2.

Therefore, 2.71g of C2H6 will react with = (2.71 x 224)/60 = 10.12g of O2.

From the above calculation, we can see that a higher mass of O2 is needed to react with 2.71g of C2H6, therefore, O2 is the limiting reactant.

Step 4:

Determination of the maximum mass of CO2 produced when 2.71 g of ethane is mixed with 7.6 g of oxygen.

The limiting reactant is used to determine the maximum mass.

From the balanced equation above,

224g of O2 produce 176g of CO2.

Therefore, 7.6g of O2 will produce = (7.6 x 176)/224 = 5.97g ≈ 6g of CO2

From the calculations made above, the maximum mass of CO2 produced is 5.97 ≈ 6g

level 4 to level 2 : ))