Question

Liquid hexane (CH_{2}, (CH_{2})_{4}CH_{3}) will react with gaseous oxygen (O_{2}) to produce gaseous carbon dioxide (CO_{2}) and gaseous water (H_{2},O). Suppose 7.76 g of hexane is mixed with 42. g of oxygen. Calculate the minimum mass of hexane that could be left over by the chemical reaction. Round your answer to 2 vignificant digits.

Answer 1

The minimum mass of hexane that could be left over by the chemical reaction can be calculated using stoichiometry and the balanced chemical equation. Based on the given values, the minimum mass of hexane left over is approximately 3.23 g.

Step-by-step explanation:

In order to answer this question, we need to balance the chemical equation for the reaction between hexane and oxygen. The balanced equation is:

2C6H14 + 19O2 → 12CO2 + 14H2O

Using the balanced equation, we can calculate the moles of hexane and oxygen. Given that the mass of hexane is 7.76 g and the mass of oxygen is 42 g, we can calculate the moles of hexane:

Moles of hexane = 7.76 g / molar mass of hexane

The minimum mass of hexane that could be left over can be calculated using the mole ratio between hexane and oxygen from the balanced equation. From the equation, we can see that 2 moles of hexane react with 19 moles of oxygen. Therefore, the mole ratio is 2:19. Using this ratio, we can calculate the moles of oxygen:

Moles of oxygen = moles of hexane * (19/2)

Finally, we can convert the moles of oxygen to grams:

Mass of oxygen = moles of oxygen * molar mass of oxygen

The minimum mass of hexane that could be left over is the difference between the initial mass of hexane and the mass of oxygen:

Minimum mass of hexane left over = initial mass of hexane - mass of oxygen

Plugging in the given values and calculating the answer, we find that the minimum mass of hexane that could be left over is approximately 3.23 g.

Answer 2

After balancing the combustion reaction of hexane and calculating the moles of reactants, it's established that hexane is the limiting reactant. Theoretically, no hexane should be left; thus, the minimum mass of hexane left over is 0 grams.

Step-by-step explanation:

To calculate the minimum mass of hexane left over after reacting with oxygen, we first need to write a balanced chemical equation for the combustion of hexane (C6H14):

2 C6H14(l) + 19 O2(g) → 12 CO2(g) + 14 H2O(g)

Using the molar mass of hexane (86.18 g/mol) and oxygen (32.00 g/mol), we convert the mass of reactants into moles:

• Moles of hexane = 7.76 g ÷ 86.18 g/mol = 0.09 mol
• Moles of oxygen = 42.0 g ÷ 32.00 g/mol = 1.31 mol

From the balanced equation, 2 moles of hexane require 19 moles of oxygen to react completely. Dividing the actual moles of each reactant by their respective stoichiometric coefficients from the balanced equation gives us:

• Hexane: 0.09 mol ÷ 2 = 0.045
• Oxygen: 1.31 mol ÷ 19 = 0.069

Oxygen has a higher ratio, meaning hexane is the limiting reactant, and no hexane should be left over theoretically. However, if excess hexane was present, the amount of hexane that could theoretically be left over would be calculated based on the stoichiometry of oxygen, the limiting reactant. Since hexane is the limiting reactant in this case, the minimum mass of hexane that could be left over is 0 grams.