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3 years ago

14

Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water. Suppose 2.1 grams of ethane i

s mixed with 3.68 grams of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction.

1 answer:

swat323 years ago

8 0

Answer:

= 3.78 g H₂O

Explanation:

2C₂H₆ + 3O₂ => 4CO₂ + 6H₂O

2.1g C₂H₆ = 2.1g/30.0 g/mol = 0.07 mole ethane

3.68g O₂ = 3.68g/32 g/mol = 0.115 mole oxygen

Limiting Reactant:

A quick way to determine limiting reactant is to divide moles of reactant by its respective coefficient in the balanced molecular equation. The smaller value is the limiting reactant.

moles ethane = 0.07 mole / 2 (the coefficient in balanced equation) = 0.035

moles oxygen = 0.115 mole / 3 (the coefficient in balanced equation) = 0.038

Since the smaller value is associated with ethane, then ethane is the limiting reactant and the problem is worked from the 0.07 moles of ethane in an excess of O₂.

From the equation stoichiometry ...

2 moles C₂H₆ in an excess of O₂ => 6 moles H₂O

then 0.07 mole C₂H₆ in an excess of O₂ => 6/2(0.07 moles H₂O = 0.21 mole

Converting to grams of water produced

= 0.21 mole H₂O X 18 g/mol = 3.78 g H₂O

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