Well, it seems like ethane and oxygen are having a hot date! Let's see how it goes.
To find out the minimum mass of ethane left over, we first need to figure out which reactant is limiting. To do that, we'll compare the moles of ethane and oxygen.
The molar mass of ethane (C2H6) is 30.07 g/mol, while the molar mass of oxygen (O2) is 32 g/mol.
Converting the mass of ethane and oxygen to moles, we get:
Moles of ethane: 17 g / 30.07 g/mol ≈ 0.564 mol
Moles of oxygen: 107 g / 32 g/mol ≈ 3.344 mol
Using the balanced equation, we see that the ratio of moles of ethane to oxygen is 1:3. So, 1 mol of ethane requires 3 mol of oxygen.
Since we have fewer moles of ethane than the stoichiometric ratio requires, we know that ethane is the limiting reagent. This means that all ethane will react, and there will be some oxygen left over.
Using the stoichiometric ratio, we can calculate the moles of oxygen required to react with all the ethane:
Moles of ethane: 0.564 mol
Moles of oxygen required: 3 mol O2/mol C2H6 x 0.564 mol C2H6 ≈ 1.692 mol O2
Now, let's convert this moles of oxygen back to grams:
Mass of oxygen required: 1.692 mol x 32 g/mol ≈ 54.144 g
Subtracting the mass of oxygen required from the initial mass of oxygen, we can find the minimum mass of ethane left over:
Mass of oxygen: 107 g
Mass of oxygen required: 54.144 g
Mass of oxygen left over: 107 g - 54.144 g ≈ 52.856 g
Therefore, the minimum mass of ethane left over is approximately 52.856 g.
Hope that helps! Keep those reactions spicy!