Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 7.82 g of ethane is mixed with 9.9 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to significant digits

4.79 g of water

Explanation:

From the reaction equation;

C2H6(g) + 7/2O2(g) > 2CO2(g) + 3H2O(g)

Next we convert the given masses of reactants to moles of reactants.

For ethane; number of moles = mass/molar mass= 7.82g/ 30gmol-1= 0.261 moles

For oxygen; number of moles= 9.9 g/32gmol-1= 0.31 moles

Next we determine the limiting reactant, the limiting reactant yields the least amount of product.

For ethane;

From the reaction equation,

1 mole of ethane yields 3 moles of water

0.261 moles of Ethan yields 0.261 ×3 = 0.783 moles of water

For oxygen;

3.5 moles of oxygen yields 3 moles of water

0.31 moles of oxygen yields 0.31 × 3/3.5 = 0.266 moles of water

Hence oxygen is the limiting reactant.

Mass of water produced = 0.266 moles of water × 18gmol-1 = 4.79 g of water

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