Question

Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Supposed 0.301 g of methane is mixed with 0.28 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits

Answer 1

Answer:

The maximum mass of CO2 that could be produced is 0.1925 grams

Explanation:

Step 1: Data given

Mass of methane = 0.301 grams

Mass of oxygen = 0.28 grams

Molar mass of ethane = 16.04 g/mol

Molar mass of oxygen = 32 g/mol

Step 2: The balanced equation

CH4 + 2O2 → CO2 + 2H2O

For 1 mole methane consumed, we need 2 moles of O2 to produce 1 mole of CO2 and 2 moles of H2O

Step 3: Calculate moles of methane

Number of moles =Mass / Molar mass

Number of moles CH4 = 0.301 grams /16.04 g/mol

Moles CH4 = 0.0188 moles

Step 4: Calculate moles oxygen

Number of moles oxygen = 0.28 grams / 32 g/mol

Moles oxygen = 0.00875 moles

Step 5: the limiting reactant

For 1 mole methane consumed, we need 2 moles of O2

Oxygen is the limiting reactant. It will be completely consumed.

Methane is in excess. There will be consumed 0.00875 /2  = 0.004375 moles

There will remain 0.0188 - 0.004375 = 0.014425 moles

Step 6: Calculate moles of CO2

For 1 mole methane consumed, we need 2 moles of O2 to produce 1 mole of CO2

For 0.00875 moles oxygen , we have 0.004375 moles CO2

Step 7: Calculate mass of CO2

Mass CO2 = moles CO2 * Molar mass CO2

Mass CO2 =0.004375 * 44.01 g/mol = 0.1925 grams CO2

The maximum mass of CO2 that could be produced is 0.1925 grams