Answer:
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.
