Liquid hexane CH3CH24CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Supposed "54.3 g" of hexane is mixed with 82. g of oxygen. Calculate the minimum mass of hexane that could be left over by the chemical reaction. Round your answer to 2 significant digits.

31.1 g of C6H14.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2C6H14 + 19O2 —> 12CO2 + 14H2O

Next, we shall determine the masses of C6H14 and O2 that reacted from the balanced equation.

This is illustrated below:

Molar mass of C6H14 = (12×6) + (14×1)

= 72 + 14

= 86 g/mol

Mass of C6H14 from the balanced equation = 2 × 86 = 172 g

Molar mass of O2 = 16×2 = 32 g/mol

Mass of O2 from the balanced equation = 19 × 32 = 608 g

Summary:

From the balanced equation above,

172 g of C6H14 reacted with 608 g of O2.

Next, we shall determine the mass of C6H14 needed to react with 82 g of O2. This can be obtained as follow:

From the balanced equation above,

172 g of C6H14 reacted with 608 g of O2.

Therefore, Xg of C6H14 will react with 82 g of O2 i.e

Xg of C6H14 = (172 × 82)/608

Xg of C6H14 = 23.2 g

Therefore, 23.2 g of C6H14 is needed to react with 82 g of O2.

Finally, we shall determine the leftover mass of hexane, C6H14. This is illustrated below:

Mass of C6H14 given = 54.3 g

Mass of C6H14 that reacted = 23.2 g

Mass of C6H14 leftover =?

Mass of C6H14 leftover = (Mass of C6H14 given) – (Mass of C6H14 that reacted)

Mass of C6H14 leftover = 54.3 – 23.2

Mass of C6H14 leftover = 31.1 g

Therefore, 31.1 g of C6H14 were leftover.

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