If 152 grams of ethane (C2H6) are reacted with 231 grams of oxygen gas, what is the excess reactant?

2C2H6 + 7O2(g) → 4CO2(g) + 6H2O(g)

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To determine the excess reactant, we need to compare the amounts of reactants used to the stoichiometry of the balanced equation.

First, we need to calculate the number of moles for each reactant using their molar masses.

The molar mass of C2H6 (ethane) is:
2 * (12.01 g/mol) + 6 * (1.01 g/mol) = 30.07 g/mol

The molar mass of O2 (oxygen gas) is:
2 * (16.00 g/mol) = 32.00 g/mol

To calculate the number of moles of each reactant, we use the formula:

moles = mass / molar mass

For ethane (C2H6):
moles of C2H6 = 152 g / 30.07 g/mol ≈ 5.05 mol

For oxygen gas (O2):
moles of O2 = 231 g / 32.00 g/mol ≈ 7.22 mol

Next, we need to determine the ratio of moles of reactants in the balanced equation to find the limiting reactant.

The balanced equation shows that 2 moles of C2H6 react with 7 moles of O2.

So, for every 2 moles of C2H6, we need 7/2 = 3.5 moles of O2.

Since we have 5.05 moles of C2H6 and 7.22 moles of O2, we can calculate the amount of O2 required based on the ratio:

required moles of O2 = (5.05 mol C2H6) * (3.5 mol O2 / 2 mol C2H6) ≈ 8.79 mol O2

Now we can compare the required moles of O2 to the actual moles of O2 used:

excess O2 = actual moles of O2 - required moles of O2
excess O2 = 7.22 mol - 8.79 mol ≈ -1.57 mol

Since the result is negative (-1.57 mol), it means that oxygen gas (O2) is the excess reactant.

To determine the excess reactant in a chemical reaction, we first need to find the mole ratio between the reactants in the balanced chemical equation. In this case, the balanced equation is:

2C2H6 + 7O2(g) → 4CO2(g) + 6H2O(g)

From the balanced equation, we can see that the ratio between ethane (C2H6) and oxygen gas (O2) is 2:7. This means that for every 2 moles of ethane, we need 7 moles of oxygen gas for the reaction to occur completely.

Now, let's calculate the number of moles of ethane and oxygen gas using their respective molar masses.

The molar mass of ethane (C2H6) is calculated as:
(2 * molar mass of carbon) + (6 * molar mass of hydrogen)

The molar mass of carbon is approximately 12.01 g/mol, and the molar mass of hydrogen is approximately 1.01 g/mol. Therefore, the molar mass of ethane is:
(2 * 12.01 g/mol) + (6 * 1.01 g/mol) = 30.07 g/mol

To determine the moles of ethane, we divide the given mass (152 g) by the molar mass (30.07 g/mol):
152 g / 30.07 g/mol = 5.06 mol of ethane

Similarly, we can calculate the moles of oxygen gas by dividing the given mass (231 g) by its molar mass (32 g/mol since each oxygen atom has a molar mass of 16 g/mol):
231 g / 32 g/mol = 7.22 mol of oxygen gas

Now, we compare the moles of ethane and oxygen gas to determine the limiting reactant and the excess reactant.

Since the mole ratio from the balanced equation tells us that 2 moles of ethane react with 7 moles of oxygen gas, we need 4/2 = 2 moles of ethane for every 7/2 = 3.5 moles of oxygen gas.

Now, let's compare the actual moles we have:
5.06 mol of ethane vs. 7.22 mol of oxygen gas.

Since we have more than enough oxygen gas (7.22 mol) compared to what is required (3.5 mol), oxygen gas is in excess. Therefore, the excess reactant in this reaction is oxygen gas.

1. mols C2H6 = grams/molar mass = ?

2. mols O2 = grams/molar mass = ?
3. Convert mols C2H6 to mols CO2 using the coefficients in the balanced equation.
4. Do the same with mols O2 to mols CO2.
5. In limiting regent problems, the SMALLER amount of CO2 produced is the correct value and the reagent producing that smaller value is the limiting reagent; the other reagent is the excess reagent.