Answer :
Final answer:
The theoretical yield of water in the reaction of 13.9 g of butane and 98.3 g of oxygen gas is calculated using the stoichiometric relationship from the balanced chemical equation and the molar masses of the compounds. In this case, the theoretical yield of water is approximately 21.5 g.
Explanation:
The reaction in question appears to be a combustion reaction where gaseous butane (C4H10) reacts with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O). The balanced equation for this reaction is:
2 C4H10 (g) + 13 O2 (g) --> 8 CO2 (g) + 10 H2O (g)
In this equation, 2 moles of butane react with 13 moles of oxygen gas to yield 8 moles of carbon dioxide and 10 moles of water. Using the concept of molar masses and the stoichiometric relationship in the balanced equation, one can calculate the theoretical yield of water.
The molar mass of butane (C4H10) is about 58.12 g/mol and that of water (H2O) is 18.02 g/mol. Thus, the moles of butane and water in 13.9 g of butane and 98.3 g of oxygen can be calculated as follows:
Moles of butane = 13.9 g ÷ 58.12 g/mol = 0.239 mol
Moles of water = 0.239 mol × (10 moles of H2O / 2 moles of C4H10) = 1.195 mol
Theoretical yield of water = 1.195 mol × 18.02 g/mol = 21.5 g
The theoretical yield of water in this reaction of 13.9 g of butane and 98.3 g of oxygen gas is therefore approximately 21.5 g.
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