For the following reaction, 5.20 grams of propane (C3H8) are allowed to react with with 23.1 grams of oxygen gas.

propane (C3H8) (g) + oxygen (g) carbon dioxide (g) + water (g)

What is the maximum amount of carbon dioxide that can be formed? grams

What is the FORMULA for the limiting reagent

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To determine the maximum amount of carbon dioxide that can be formed, we need to find the limiting reagent in the reaction. The limiting reagent is the reactant that is completely consumed and determines the amount of product that can be formed.

To find the limiting reagent, we need to calculate the number of moles of each reactant.

1. Convert the mass of propane (C3H8) to moles:
Molar mass of propane (C3H8) = 3(12.01 g/mol) + 8(1.01 g/mol) = 44.11 g/mol

Number of moles of propane = (mass of propane) / (molar mass of propane)
= 5.20 g / 44.11 g/mol
= 0.118 mol

2. Convert the mass of oxygen gas (O2) to moles:
Molar mass of oxygen (O2) = 2(16.00 g/mol) = 32.00 g/mol

Number of moles of oxygen = (mass of oxygen) / (molar mass of oxygen)
= 23.1 g / 32.00 g/mol
= 0.723 mol

3. Determine the mole ratio between propane and carbon dioxide in the balanced equation:
From the balanced equation: 1 mol of propane reacts to produce 3 mol of carbon dioxide.

4. Calculate the number of moles of carbon dioxide that can be formed:
Number of moles of carbon dioxide = (number of moles of propane) * (mole ratio)
= 0.118 mol * (3 mol C3H8 / 1 mol CO2)
= 0.354 mol

5. Convert the number of moles of carbon dioxide to grams:
Molar mass of carbon dioxide (CO2) = 12.01 g/mol + 2(16.00 g/mol) = 44.01 g/mol

Mass of carbon dioxide = (number of moles of carbon dioxide) * (molar mass of carbon dioxide)
= 0.354 mol * 44.01 g/mol
= 15.58 g

Therefore, the maximum amount of carbon dioxide that can be formed is 15.58 grams.

To determine the limiting reagent, we compare the number of moles of each reactant to the stoichiometric ratio in the balanced equation. The reactant that produces the lesser amount of product is the limiting reagent.

For propane (C3H8):
Number of moles = 0.118 mol

For oxygen (O2):
Number of moles = 0.723 mol

From the balanced equation, we can see that the stoichiometric ratio between propane and carbon dioxide is 1:3. Therefore, 1 mole of propane produces 3 moles of carbon dioxide.

For propane (C3H8):
Number of moles of carbon dioxide produced = 0.118 mol * (3 mol CO2 / 1 mol C3H8) = 0.354 mol

For oxygen (O2):
Number of moles of carbon dioxide produced = 0.723 mol * (3 mol CO2 / 1 mol O2) = 2.169 mol

Based on the calculations, we can see that the amount of carbon dioxide produced from propane is 0.354 mol, while the amount produced from oxygen is 2.169 mol. Since the amount produced from propane is lesser, it implies that propane is the limiting reagent.

Therefore, the formula for the limiting reagent is propane (C3H8).

To determine the maximum amount of carbon dioxide that can be formed and the limiting reagent in the reaction, you need to follow these steps:

Step 1: Convert the given masses of propane (C3H8) and oxygen gas (O2) into moles.

1 mole of propane (C3H8) has a molar mass of 44.1 grams.
Therefore, (5.20 grams C3H8) x (1 mole C3H8/44.1 grams C3H8) = 0.118 moles C3H8 (rounded to 3 decimal places).

1 mole of oxygen (O2) has a molar mass of 32.0 grams.
Therefore, (23.1 grams O2) x (1 mole O2/32.0 grams O2) = 0.722 moles O2 (rounded to 3 decimal places).

Step 2: Write a balanced chemical equation for the reaction.

The balanced chemical equation for the reaction is:
C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O (g)

Step 3: Determine the stoichiometric ratio between the reactants and products.

From the balanced equation, we can see that the stoichiometric ratio between propane (C3H8) and carbon dioxide (CO2) is 1:3. This means that for every mole of propane, 3 moles of carbon dioxide are produced.

Step 4: Determine the limiting reagent.

The limiting reagent is the reactant that is completely consumed and determines the maximum amount of product that can be formed. To find the limiting reagent, compare the number of moles of each reactant to the stoichiometric ratio.

For propane (C3H8), you have 0.118 moles.
For oxygen gas (O2), you have 0.722 moles.

The stoichiometric ratio between propane and oxygen is 1:5. Therefore, to react completely, 0.118 moles of propane would require 0.118 moles x 5 = 0.590 moles of oxygen.

Since you have 0.722 moles of oxygen, it is in excess compared to the required amount. Therefore, propane is the limiting reagent.

Step 5: Calculate the maximum amount of carbon dioxide that can be formed.

Since propane is the limiting reagent, we can calculate the maximum moles of carbon dioxide produced based on this.

The stoichiometric ratio between propane and carbon dioxide is 1:3. Therefore, the maximum moles of carbon dioxide produced would be 0.118 moles (moles of propane) x 3 (stoichiometric ratio) = 0.354 moles.

Step 6: Convert the moles of carbon dioxide to grams.

1 mole of carbon dioxide (CO2) has a molar mass of 44.0 grams.
Therefore, (0.354 moles CO2) x (44.0 grams CO2/1 mole CO2) = 15.5 grams of carbon dioxide.

So, the maximum amount of carbon dioxide that can be formed is 15.5 grams.

The formula for the limiting reagent is C3H8 (propane).