Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 7.22 g of ethane is mixed with 15. g of oxygen. Calculate the minimum mass of ethane that could be left over by the chemical reaction. Round your answer to significant digits.

The remaining mass of ethane is 3.19 grams

Explanation:

Step 1: Data given

Mass of ethane = 7.22 grams

Mass of oxygen = 15.0 grams

Molar mass ethane = 30.07 g/mol

Molar mass oxygen = 32.0 g/mol

Step 2: The balanced equation

2C2H6 + 7O2 → 4CO2 + 6H2O

Step 3: Calculate moles ethane

Moles ethane = 7.22 grams / 30.07 g/mol

Moles ethane = 0.240 moles

Step 4: Calculate moles oxygen

Moles oxygen = 15.0 grams / 32.0 g/mol

Moles oxygen = 0.46875 moles

Step 5: Calculate limiting reactant

For 2 moles ethane we need 7 moles oxygen to produce 4 moles CO2 and 6 moles H2O

Oxygen is the limiting reactant. It will completely be consumed (0.46875 moles). Ethane is in excess. There will react 0.46875 / 3.5 = 0.134 moles

There will remain 0.240 - 0.134 = 0.106 moles

Step 6: Calculate mass of remaining ethane

Mass ethane = 0.106 moles * 30.06 g/mol

Mass ethane = 3.19 grams

The remaining mass of ethane is 3.19 grams

the old alchemists made significant contributions to science even though there starting premises were off a bit. their conclusions after failure would steer them into better directions for investigation. science is no more than close observation and the initial mistakes result in rethinking and retrying experiments.

b) mutualism

hope this : )