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