Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 17. g of ethane is mixed with 107. g of oxygen. Calculate the minimum mass of ethane that could be left over by the chemical reaction. Be sure your answer has the correct number of significant digits.
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!
To calculate the minimum mass of ethane that could be left over, we need to determine which reactant is in excess.
1. Write the balanced equation for the reaction:
C2H6 + O2 -> CO2 + H2O
2. Calculate the molar mass of ethane (C2H6):
C = 12.01 g/mol
H = 1.008 g/mol
Molar mass of ethane (C2H6) = (2 * C) + (6 * H)
= (2 * 12.01 g/mol) + (6 * 1.008 g/mol)
= 30.07 g/mol
3. Calculate the molar mass of oxygen (O2):
O = 16.00 g/mol
Molar mass of oxygen (O2) = 2 * O
= 2 * 16.00 g/mol
= 32.00 g/mol
4. Find the number of moles for each reactant:
Number of moles of ethane = mass of ethane / molar mass of ethane
= 17.0 g / 30.07 g/mol
= 0.5653 mol (rounded to 4 decimal places)
Number of moles of oxygen = mass of oxygen / molar mass of oxygen
= 107.0 g / 32.00 g/mol
= 3.34375 mol (rounded to 5 decimal places)
5. Determine the limiting reactant:
To find the limiting reactant, compare the mole ratio of ethane to oxygen based on the balanced equation:
2 moles of ethane react with 7 moles of oxygen.
First, convert the moles of ethane to moles of oxygen:
0.5653 mol ethane * (7 mol oxygen / 2 mol ethane)
= 1.97955 mol oxygen (rounded to 5 decimal places)
We see that the number of moles of oxygen is greater than the calculated 1.97955 mol, which means oxygen is in excess and ethane is the limiting reactant.
6. Calculate the mass of ethane used:
mass of ethane used = moles of ethane * molar mass of ethane
= 0.5653 mol * 30.07 g/mol
= 16.9923 g (rounded to 4 decimal places)
7. Calculate the mass of ethane left over:
mass of ethane left over = mass of ethane initially - mass of ethane used
= 17.0 g - 16.9923 g
= 0.0077 g (rounded to 3 decimal places)
Therefore, the minimum mass of ethane that could be left over is 0.0077 g.
To calculate the minimum mass of ethane that could be left over by the chemical reaction between ethane and oxygen, we need to determine the limiting reactant. The limiting reactant is the reactant present in the smallest stoichiometric amount, which will determine the maximum amount of product that can be formed.
1. Start by writing the balanced chemical equation for the reaction:
C2H6 + O2 → CO2 + H2O
2. Calculate the molar masses of ethane (C2H6) and oxygen (O2):
Molar mass of C2H6 = 2(12.01 g/mol) + 6(1.01 g/mol) = 30.07 g/mol
Molar mass of O2 = 2(16.00 g/mol) = 32.00 g/mol
3. Convert the given masses of ethane and oxygen to moles:
Moles of ethane = 17. g / 30.07 g/mol = 0.565 mol (to three significant digits)
Moles of oxygen = 107. g / 32.00 g/mol = 3.34 mol (to three significant digits)
4. Determine the stoichiometric ratio between ethane and oxygen:
From the balanced chemical equation, the ratio is 1:1. This means that 1 mole of ethane reacts with 1 mole of oxygen.
5. Compare the moles of ethane and oxygen to find the limiting reactant:
Since the ratio is 1:1, we can see that we have an excess of O2. Therefore, ethane is the limiting reactant.
6. Calculate the amount of ethane left over:
Using the stoichiometric ratio, 0.565 mol of ethane reacts completely with 0.565 mol of oxygen. Since ethane is the limiting reactant, there will be no oxygen left over. Therefore, the amount of ethane left over is 0 mol.
7. Convert the moles of remaining ethane to grams:
Mass of ethane left over = 0 mol × 30.07 g/mol = 0 g
Therefore, the minimum mass of ethane that could be left over by the chemical reaction is 0 grams.
2C2H6 + 7O2 ==> 4CO2 + 6H2O
mols ethane = grams/molar mass = approx 0.0.57 but you should confirm this and all of the others that follow because I estimate.
mols O2 = 107/32 = 3.34
Now convert mols ethane to mols CO2. That's 0.57 x (4 mols CO2/2 mols C2H6) = 0.57 x 2 = 1.13
Do the same for O2. That's 3.34 x (4 mols CO2/7 mols O2) = 1.9
I have news for you. The limiting reagent is ethane so none of it will be left over. 0.00 is the correct answer although the problem makes it sound as if O2 is the limiting reagent.