Gaseous butane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water. Suppose 25. g of butane is mixed with 43.8 g of oxygen. Calculate the minimum mass of butane that could be left over by the chemical reaction. Round your answer to significant digits.

Question

Caroline West

Chemistry

7 February, 08:21

Gaseous butane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water. Suppose 25. g of butane is mixed with 43.8 g of oxygen. Calculate the minimum mass of butane that could be left over by the chemical reaction. Round your answer to significant digits.

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Answers (1)

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Huang · Answer 1

13g

Explanation:

Step 1:

The equation for the reaction. This is given below:

C4H10 + O2 - > CO2 + H2O

Step 2:

Balancing the equation. The equation can be balanced as follow:

C4H10 + O2 - > CO2 + H2O

There are 4 atoms of C on left side and 1 atom on the right side. It can be balance by putting 4 in front of CO2 as shown below:

C4H10 + O2 - > 4CO2 + H2O

There are 10 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 5 in front of H2O as shown below:

C4H10 + O2 - > 4CO2 + 5H2O

There are a total of 13 atoms of O on the right side and 2 atoms on the left side. It can be balance by putting 13/2 in front of O2 as shown below:

C4H10 + 13/2O2 - > 4CO2 + 5H2O

Multiply through by 2 to clear the fraction

2C4H10 + 13O2 - > 8CO2 + 10H2O

Now the equation is balanced.

Step 3:

Determination of the masses of C4H10 and O2 that reacted from the balanced equation. This is illustrated below:

2C4H10 + 13O2 - > 8CO2 + 10H2O

Molar Mass of C4H10 = (12x4) + (10x1) = 48 + 10 = 58g/mol

Mass of C4H10 from the balanced equation = 2 x 58 = 116g

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

Mass of O2 from the balanced equation = 13 x 32 = 416g

From the balanced equation above, 116g of C4H10 reacted will 416g of O2.

Step 4:

Determination of the mass of C4H10 that will react with 43.8 g of O2. This is illustrated below:

From the balanced equation above, 116g of C4H10 reacted will 416g of O2.

Therefore, Xg of C4H10 will react with 43.8g of O2 i. e

Xg of C4H10 = (116 x 43.8) / 416

Xg of C4H10 = 12g

Therefore, 12g of C4H10 reacted with 43.8g of O2.

Step 5:

Determination of the leftover Mass of C4H10. This is illustrated below:

Mass of C4H10 given = 25g

Mass of C4H10 that reacted = 12g

Mass of C4H10 remaining = ?

Mass of C4H10 remaining = (Mass of C4H10 given) - (Mass of C4H10 that reacted)

Mass of C4H10 remaining = 25 - 12

Mass of C4H10 remaining = 13g