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

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

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