Quiz 2 - CEGR 338

The general formula for alkanes is CₙH₂ₙ₊₂. What does n represent in this formula?

• The number of hydrogen atoms in the molecule

• The number of carbon atoms in the molecule

• The molecular mass of the alkane

• The number of bonds in the molecule

If N = 5, which corresponds to pentane, what is the molecular formula of pentane?

• C₅H₈

• C₅H₁₀

• C₅H₁₂

• C₅H₁₄

Balance the following combustion reaction:

C₆H₁₄+O₂→CO₂+H₂O

• 2C_6​H_14​+19O_2​ → 6CO_2​+14H_2​O

• 2C_6​H_14​+24O_2​ → 12CO_2​+7H_2​O

• 2C_6​H_14​+19O_2​ → 12CO_2​+14H_2​O

• 2C_6​H_14​+19O_2​ → 6CO_2​+7H_2​O

Calculate the molar mass of hexane (C₆H₁₄)

Molar mass:

• Carbon (C) = 12.01 g/mol

• Hydrogen (H) = 1.008 g/mol

• 72.06 g/mol

• 86.17 g/mol

• 100.12 g/mol

• 58.12 g/mol

Convert 100 g of hexane (C₆H₁₄) to moles.
(Molar mass of hexane = 86.17 g/mol)

• 0.86 moles

• 1.16 moles

• 1.39 moles

• 1.92 moles

Using the balanced equation you balanced before:

2 C₆H₁₄ + 19 O₂ → 12 CO₂ + 14 H₂O

How many moles of oxygen (O₂) are required to burn 1.16 moles of hexane (C₆H₁₄)?

• 9.50 moles O₂

• 10.25 moles O₂

• 11.03 moles O₂

• 12.50 moles O₂

Convert 11.03 moles of oxygen (O₂) to grams. The molar mass of O₂ is 32.00 g/mol.

• 321.0 g

• 353.0 g

• 288.5 g

• 375.2 g

Using the previously calculated 11.03 moles of O₂, how many liters of oxygen gas (O₂) is required to burn 100g of hexane (C₆H₁₄) at standard temperature and pressure (STP)?
(1 mole of gas = 22.4 L at STP)

• 123.5 L

• 220.6 L

• 247.0 L

• 275.8 L

 If the volume of oxygen (O₂) required to burn 100g of hexane is 247.0 L, and assuming air is 21% oxygen and , how many liters of air are needed?

•  1176.2 L

•  950.6 L

• 576.6 L

• 220.2 L

Knowing that 100g of hexane (C₆H₁₄) represents 1.16 moles of hexane, how many moles of CO₂ are produced by burning 1.16 moles of hexane?

• 7.0 moles of CO₂

• 12.0 moles of CO₂

• 1.16 moles of CO₂

• 19.0 moles of CO₂