Consider the balanced chemical equation: A + 5B → 3C + 4D. Determine the limiting reactant when equal masses of A and B are reacted. Also, identify compound B and explain your answer. Additionally, find the empirical and molecular formulas of compound A, which is a hydrocarbon with a carbon mass percentage of 81.71%.
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To determine the limiting reactant, we compare the molar masses of A and B. If the molar mass of A is greater than that of B, then B is the limiting reactant because it is consumed at a higher rate. Conversely, if the molar mass of B is greater than that of A, then A is the limiting reactant because it is consumed at a lower rate. In this case, since compound B is a diatomic molecule and commonly involved in combustion reactions with hydrocarbons to produce carbon dioxide and water, the most likely candidate for compound B is molecular oxygen (O2).
To find the empirical formula of compound A, we calculate the ratio of carbon to hydrogen. Given that the compound is 81.71% carbon by mass, it is 18.29% hydrogen by mass. Assuming we have 100 g of the compound, we have 81.71 g of carbon and 18.29 g of hydrogen. Converting these masses to moles, we find that there are approximately 6.808 moles of carbon and 18.15 moles of hydrogen. The ratio of moles of carbon to moles of hydrogen is approximately 1:3, leading to the empirical formula CH3.
Since compound A has the same molar mass as carbon dioxide, we calculate the molar mass of the empirical formula CH3, which is 15.03 g/mol. By dividing the molar mass of A (44.01 g/mol) by the molar mass of CH3, we find a ratio of approximately 3. This indicates that the molecular formula of compound A is 3 times the empirical formula, resulting in C3H9. However, this formula is not common for hydrocarbons. The closest common hydrocarbon is C3H8, which is propane. It is possible that the given percentage of carbon by mass (81.71%) was slightly off, leading to this discrepancy.
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