Liquid hexane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water. Suppose 4.3 g of hexane is mixed with 7.14 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to significant digits.

We can produce 6.20 grams of CO2

Explanation:

Step 1: Data given

Mass of hexane = 4.3 grams

Molar mass of hexane = 86.18 g/mol

Mass of oxygen = 7.14 grams

Molar mass of oxygen = 32.0 g/mol

Step 2: The balanced equation

2C6H14 + 19O2 → 12CO2 + 14H2O

Step 3: Calculate moles

Moles = mass / molar mass

Moles hexane = 4.3 grams / 86.18 g/mol

Moles hexane = 0.0499 moles

Moles oxygen = 7.14 grams / 32.0 g/mol

Moles oxygen = 0.2231 moles

Step 4: Calculate the limiting reactant

For 2 moles hexane we need 19 moles O2 to produce 12 moles CO2 and 14 moles H2O

Oxygen is the limiting reactant. It will completely be consumed (0.2231 moles). Hexane is in excess. There will react 2/19 * 0.2231 = 0.02348 moles

There will be porduced 12/19 * 0.2231 = 0.1409 moles CO2

Step 5: Calculate mass CO2

Mass CO2 = moles CO2 * molar mass CO2

Mass CO2 = 0.1409 moles * 44.01 g/mol

Mass CO2 = 6.20 grams

We can produce 6.20 grams of CO2