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.

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.

it should have an internal chamber and two thermometers sticking out of the top. the internal chamber will also be closed off from the outside.