Question
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.
<b>These are stoichiometry problems. There are four steps to most. Here is how to do the firt one in detail.</b>
1.) How many grams of 2-propanol are expected from the complete reaction of 672 g of propene?
<b>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</b>
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.
<b>These are stoichiometry problems. There are four steps to most. Here is how to do the firt one in detail.</b>
1.) How many grams of 2-propanol are expected from the complete reaction of 672 g of propene?
<b>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</b>