Question

Suppose 9.1 g of octane is mixed with 59.5 g of oxygen. Calculate the minimum mass of octane that could be leftover by the chemical reaction

Answer 1

To calculate the minimum mass of octane leftover, we need to determine the limiting reagent, convert the given masses to moles, and use stoichiometry to calculate the minimum mass of octane leftover. In this case, oxygen is the limiting reagent and all of it will be consumed, resulting in zero leftover octane.

Step-by-step explanation:

To find the minimum mass of octane leftover, we need to determine the limiting reagent, which is the reactant that will be completely consumed in the chemical reaction. First, we need to convert the given masses of octane and oxygen into moles using their molar masses (C8H18 = 114 g/mol, O2 = 32 g/mol). Once we have the number of moles of each reactant, we can compare the moles using the balanced chemical equation to determine the limiting reagent.

From the balanced equation:

C8H18 + 12.5O2 → 8CO2 + 9H2O

By dividing the number of moles of octane by the stoichiometric coefficient of octane in the balanced equation (8), and the number of moles of oxygen by the stoichiometric coefficient of oxygen in the balanced equation (12.5), we can see that oxygen is the limiting reagent. This means that all of the oxygen will be consumed in the reaction, leaving some octane as leftover.

To calculate the minimum mass of octane leftover, we can use the following steps:

1. Convert the number of moles of oxygen to moles of octane using the mole ratio from the balanced equation (12.5 moles of O2 : 8 moles of octane).
2. Convert the moles of octane to grams using the molar mass of octane (114 g/mol).
3. Subtract the resulting mass of octane from the initial mass of octane to find the minimum mass leftover.

Let's calculate the minimum mass of octane leftover:

1. Convert the mass of oxygen to moles: 59.5 g O2 * (1 mol O2/32 g O2) = 1.859 mol O2
2. Convert moles of O2 to moles of octane using the mole ratio: 1.859 mol O2 * (8 mol octane/12.5 mol O2) = 1.190 mol octane
3. Convert moles of octane to grams: 1.190 mol octane * (114 g octane/1 mol octane) = 135.66 g octane
4. Subtract the mass of octane from the initial mass: 9.1 g octane - 135.66 g octane = -126.56 g octane

The minimum mass of octane leftover is -126.56 g. However, since we cannot have a negative mass, the actual amount of octane leftover in this reaction would be zero.