Question

Liquid hexane (CH3(CH2)4CH3) reacts with gaseous oxygen gas (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O). What is the theoretical yield of water formed from the reaction of 62.9 g of hexane and 158. g of oxygen gas?

Be sure your answer has the correct number of significant digits in it.

Answer 1

The theoretical yield of water from the reaction of 62.9 g of hexane and 158 g of oxygen gas is 26.1 g, considering oxygen as the limiting reactant. The stoichiometry of the balanced reaction and mole-to-mass conversions are used for this calculation.

Step-by-step explanation:

The question asks for the theoretical yield of water (H2O) when liquid hexane (CH3(CH2)4CH3) reacts with gaseous oxygen (O2) to produce CO2 and H2O. To calculate this, we need the balanced chemical equation for the combustion of hexane, which is:

C6H14(l) + 19/2 O2(g) → 6 CO2(g) + 7 H2O(g)

This equation indicates that one mole of hexane reacts with 19/2 moles of oxygen to produce 6 moles of CO2 and 7 moles of H2O. Using stoichiometry, we first determine the moles of hexane and oxygen, then find the limiting reactant, and finally calculate the theoretical yield of water based on the limiting reactant.

First, calculate the moles of hexane:

• Molar mass of hexane (C6H14) = 86.18 g/mol
• Moles of hexane (62.9 g / 86.18 g/mol) = 0.730 moles

Then, calculate the moles of oxygen:

• Molar mass of O2 = 32.00 g/mol
• Moles of oxygen (158 g / 32.00 g/mol) = 4.9375 moles

Next, using the stoichiometry of the balanced equation, calculate the number of moles of water produced from the moles of hexane:

• (0.730 moles of hexane) x (7 moles of water / 1 mole of hexane) = 5.11 moles of water

Now, determine if oxygen is the limiting reactant:

• (4.9375 moles of oxygen) x (7 moles of water / 19/2 moles of oxygen) = 1.446 moles of water

Since the theoretical yield of water based on oxygen (1.446 moles) is less than that based on hexane (5.11 moles), oxygen is the limiting reactant. Therefore, the theoretical yield of water is 1.446 moles.

Finally, calculate the mass of this theoretical yield:

• (1.446 moles of water) x (18.02 g/mol) = 26.06 g of water

So, the theoretical yield of water from the reaction of 62.9 g of hexane and 158 g of oxygen gas is 26.06 g, and should be reported with three significant figures as 26.1 g of water.

Answer 2

64.8g of water.

Step-by-step explanation:

First we must give the balanced reaction equation.

C6H14 + 19/2 O2 ===> 6CO2 + 7H2O ...

moles of hexane present = 62.9g x 1 mole/86 g = 0.73 moles

moles O2 present = 158 g x 1 mole/32g = 4.9 moles

Hence ; oxygen is limiting...

moles of H2O that can be produced

If 9.5 moles of oxygen yields 7 moles of water

4.9 moles of oxygen yields 7×4.9/9.5 = 3.6 moles

Mass of water= 3.6×18= 64.8g of water.