If 20.0 grams of propane are burned with 50.0 grams of oxygen:

What mass of CO₂ is produced?
What mass of water is produced?
What mass of which reactant was excess?

Thank you so much.

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To find the mass of CO₂ produced, we need to use the balanced equation for the combustion of propane:

C₃H₈ + 5O₂ -> 3CO₂ + 4H₂O

First, we need to determine the limiting reactant. The limiting reactant is the one that is completely consumed, thus limiting the amount of product that can be formed.

Let's calculate the number of moles of propane and oxygen:

Molar mass of propane (C₃H₈) = 44.1 g/mol
Molar mass of oxygen (O₂) = 32.0 g/mol

Number of moles of propane = mass of propane / molar mass of propane
Number of moles of propane = 20.0 g / 44.1 g/mol

Number of moles of oxygen = mass of oxygen / molar mass of oxygen
Number of moles of oxygen = 50.0 g / 32.0 g/mol

Now, we need to compare the number of moles of propane and oxygen using the balanced equation to know which one is the limiting reactant.

The balanced equation shows that 1 mole of propane reacts with 5 moles of oxygen. Therefore, the number of moles of oxygen needed to react with all 20.0 grams of propane can be calculated as follows:

Number of moles of oxygen needed = (number of moles of propane) x 5

Now, if the number of moles of oxygen available is greater than the number needed, then oxygen is in excess. However, if the number of moles of oxygen available is less than the number needed, then propane is in excess.

Now we can calculate the moles of propane and oxygen:

Moles of propane = 20.0 g / 44.1 g/mol
Moles of oxygen = 50.0 g / 32.0 g/mol

Number of moles of oxygen needed = (moles of propane) x 5

If the number of moles of oxygen is greater than the number of moles of oxygen needed, then oxygen is in excess. If the number of moles of oxygen is less than the number of moles of oxygen needed, then propane is in excess.

To find the mass of CO₂ produced, we can use the balanced equation:

1 mole of propane produces 3 moles of CO₂

So, the moles of CO₂ produced can be calculated as follows:

moles of CO₂ produced = (moles of propane) x 3

To find the mass of CO₂ produced, we can use the molar mass of CO₂, which is 44.0 g/mol:

mass of CO₂ produced = (moles of CO₂ produced) x molar mass of CO₂

Similarly, to find the mass of water produced, we can use the balanced equation:

1 mole of propane produces 4 moles of water

So, the moles of water produced can be calculated as follows:

moles of water produced = (moles of propane) x 4

To find the mass of water produced, we can use the molar mass of water, which is 18.0 g/mol:

mass of water produced = (moles of water produced) x molar mass of water

Once you have calculated the mass of CO₂ and water produced, you can compare it to the masses of propane and oxygen used to determine which reactant was in excess. If the mass of CO₂ or water is smaller than what should have been produced based on the stoichiometry, then the reactant that produced the smaller mass is in excess.