How many grams of O2(g) are needed to completely burn 24.0 g of C3H8(g)?

Here are the steps to work any simple stoichiometry problem (like this one). You must make a slight adjustment when you get to limiting reagent problems but they are almost the same. Print this out. Keep it. Memorize the steps.

Step 1. Write and balance the equation.
C3H8 + 5O2 ==> 3CO2 + 4H2O

Step 2. Convert grams of what you have (in this case g C3H8) to mols. mols C3H8 = grams/molar mass = 24.0/44 = 0.545

Step 3. Using the coefficient in the balanced equation, convert mols of what you have (C3H8) to mols of what you want (in this case mols O2).
0.545 mols C3H8 x (5 mols O2/1 mol C3H8 = 0.545 x (5/1) = 2.73

Step 4. Convert mols of what you want (O2) to grams. g = mols x molar mass
g = 2.727 x 32 = ? and round to 3 significant figures.

Balance the equation

C3H8 + _n_ O2 = 3 CO2 + 4 H2O

mass(O2) = n * mass(C3H8) * molarmass(O2) / molarmass(C3H8)

To answer this question, we first need to balance the combustion equation for the reaction between propane (C3H8) and oxygen (O2). The balanced equation is as follows:

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

From the balanced equation, we can see that for every 1 mole of C3H8, we need 5 moles of O2. To calculate the number of moles of C3H8, we can use the molar mass of propane.

The molar mass of C3H8 is calculated as follows:
Molar mass (C3H8) = (3 x molar mass of carbon) + (8 x molar mass of hydrogen)
= (3 x 12.01 g/mol) + (8 x 1.01 g/mol)
= 36.03 g/mol + 8.08 g/mol
= 44.11 g/mol

Now, we can calculate the number of moles of C3H8 by dividing the given mass by the molar mass:
Number of moles of C3H8 = 24.0 g / 44.11 g/mol
≈ 0.545 mol

Since the molar ratio between C3H8 and O2 is 1:5, we can calculate the number of moles of O2 using the following equation:

Number of moles of O2 = Number of moles of C3H8 x (5 moles of O2/1 mole of C3H8)
= 0.545 mol x 5
= 2.725 mol

Finally, to convert the number of moles of O2 to grams, we multiply by the molar mass of O2:

Molar mass (O2) ≈ 2 x molar mass of oxygen
≈ 2 x 16.00 g/mol
≈ 32.00 g/mol

Grams of O2 = Number of moles of O2 x molar mass of O2
= 2.725 mol x 32.00 g/mol
≈ 87.2 g

Therefore, approximately 87.2 grams of O2 are needed to completely burn 24.0 grams of C3H8.

To determine the number of grams of O2(g) needed to completely burn 24.0 g of C3H8(g), we need to use balanced chemical equations and stoichiometry.

Step 1: Write the balanced equation for the combustion of C3H8.

C3H8 + 5O2 -> 3CO2 + 4H2O

This equation tells us that for every one mole of C3H8, 5 moles of O2 are needed to completely burn it.

Step 2: Calculate the molar mass of C3H8.

C3H8: C - 3 atoms x 12.01 g/mol = 36.03 g/mol
H - 8 atoms x 1.01 g/mol = 8.08 g/mol

Molar mass of C3H8 = 36.03 g/mol + 8.08 g/mol = 44.11 g/mol

Step 3: Calculate the number of moles of C3H8.

Moles of C3H8 = mass of C3H8 / molar mass of C3H8
= 24.0 g / 44.11 g/mol
= 0.544 mol

Step 4: Use the stoichiometry of the balanced equation to find the moles of O2 required.

From the balanced equation, we can see that 1 mole of C3H8 reacts with 5 moles of O2. Therefore, the moles of O2 needed is:

Moles of O2 = moles of C3H8 x (5 moles of O2 / 1 mole of C3H8)
= 0.544 mol x (5 mol O2 / 1 mol C3H8)
= 2.72 mol

Step 5: Convert moles of O2 to grams.

Moles of O2 = 2.72 mol
Molar mass of O2 = 32.00 g/mol (approximate)

Mass of O2 = moles of O2 x molar mass of O2
= 2.72 mol x 32.00 g/mol
= 87.04 g

Therefore, approximately 87.04 grams of O2(g) are needed to completely burn 24.0 grams of C3H8(g).