How many moles of propane are expected from the complete reaction of 82.8 g of propene?

How many grams of propane are expected from the complete reaction of 101 g of propene?

Kyle-- Here is a problem I worked last night for Kira (about half way down the page). It should give you the procedure for doing these. Most of these problems are about the same.

These are stoichiometry problems. There are four steps to most. Here is how to do the firt one in detail.

1.) How many grams of 2-propanol are expected from the complete reaction of 672 g of propene?

Step 1. Write the balanced chemical equation.
propene + HOH ==> propanol
C3H6 + HOH ==> C3H8O

Step 2.
Convert what you have to mols.
mols = grams/molar mass = 672g/42.08 = 15.97 mols propene.

Step 3. Using the equation from above, use the coefficients to convert what you have (mols propene) to mols of what you want (mols 2-propanol).

mols 2-propanol = mols propene x (1 mol 2-propanol/1 mol propene).
mols 2-propanol = 15.97x(1 mol 2-propanol/1 mol propene) = 15.97 x 1/1 = 15.97 mols 2-propanol. Notice the the factor is placed so that the units of propene cancel and leave units of 2-propanol.

Step 4. Convert mols to grams.
mols 2-propanol x molar mass 2-propanol = grams 2-propanol.
15.97 x 60.1 g/mol = 959.8 g which rounds to 960 to three significant figures. 960 g 2-propanol. That is the theoratical yield.

If you get stuck on your problem please post your work and tell us what you don't understand about the next step. I hope this helps.

You don't say what reacts with the propene. But you want to convert propene, which is C3H6 to propane, which is C3H8. The ratio, however, is still 1 mol propene will produce 1 mol propane.
C3H6 + H2 ==> C3H8

To find the number of moles of propane expected from the complete reaction of 82.8 g of propene, you can follow these steps:

Step 1: Write the balanced chemical equation for the reaction:
C3H6 + H2 → C3H8

Step 2: Convert the mass of propene to moles using the molar mass of propene:
Molar mass of C3H6 = 42.08 g/mol
Moles of propene = 82.8 g / 42.08 g/mol = 1.9687 mol

Step 3: Apply the stoichiometric coefficients from the balanced equation to convert moles of propene to moles of propane:
1 mol propene produces 1 mol propane

Therefore, the number of moles of propane expected from the reaction of 82.8 g of propene is also 1.9687 mol.

To find the grams of propane expected from the complete reaction of 101 g of propene, you can follow similar steps:

Step 1: Write the balanced chemical equation for the reaction:
C3H6 + H2 → C3H8

Step 2: Convert the mass of propene to moles using the molar mass of propene:
Molar mass of C3H6 = 42.08 g/mol
Moles of propene = 101 g / 42.08 g/mol = 2.4030 mol

Step 3: Apply the stoichiometric coefficients from the balanced equation to convert moles of propene to moles of propane:
1 mol propene produces 1 mol propane

Therefore, the number of moles of propane expected from the reaction of 101 g of propene is also 2.4030 mol.

Step 4: Convert the moles of propane to grams using the molar mass of propane:
Molar mass of C3H8 = 44.10 g/mol
Grams of propane = 2.4030 mol * 44.10 g/mol = 105.8493 g

Therefore, the grams of propane expected from the reaction of 101 g of propene is approximately 105.8493 g.