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Liquid octane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 10.3 g of octane is mixed with 23. g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to significant digits.

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Answer:

9.36 g

Explanation:

The equation of the reaction is;

C8H18(g) + 25/2 O2(g) ----> 8CO2(g) + 9H2O(g)

Number of moles of octane = 10.3g/ 114 g/mol = 0.09 moles

1 mole of octane yields 9 moles of water

0.09 moles of octane yields 0.09 × 9/1 = 0.81 moles of water

Number of moles of oxygen = 23g/32g/mol = 0.72 moles

12.5 moles of oxygen yields 9 moles of water

0.72 moles of oxygen yields 0.72 × 9/12.5 = 0.52 moles of water

Hence oxygen is the limiting reactant;

Maximum mass of water produced = 0.52 moles of water × 18 g/mol = 9.36 g

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