Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water. Suppose 9.02 g of ethane is mixed with 22. g of oxygen. Calculate the minimum mass of ethane that could be left over by the chemical reaction. Round your answer to significant digits.

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