Chemistry

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

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

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

4.)Consider the reaction: KOH(s) + CO2(g) KHCO3(s). How many grams of KOH are required to react completely with 5.78 mol of CO2?

5.)Consider the reaction: 2KClO3(s) 2KCl(s) + 3O2(g). How many grams of KCl are produced from the complete reaction of 50.1 g of KClO3?

6.) What is the theoretical yield (in grams) of P4O10 if 20.5 g of P4 are reacted with 8.95 g of O2?

7.)What is the theoretical yield (in grams) of HgO if 213 g of Hg are reacted with 213 g of O2?

8.)What is the theoretical yield of aniline (in grams), beginning with 623 g of nitrobenzene? (Assume that nitrobenzene is the limiting reactant.)

9.)Based on the theoretical yield from part a, what is the percent yield if 74.9 g of aniline are obtained?

10.)What is the theoretical yield of ethane (in moles) if 9.0 mol of ethylene are reacted with 3.0 mol of H2?

11.)What is the theoretical yield of ethane (in grams) if 50.6 g of ethylene are reacted with 32.0 g of H2?

12.)What theoretical yield of formaldehyde (in grams) is expected from the reaction of 44.3 g of methanol? (Assume that methanol is the limiting reactant.)


I don't necessarily need answers as much as I need to know how to work these out. Please please please


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 equatiion.
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) = 15.97/1/1 = 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.

We shall be happy to check your answers to the remainder. If you have trouble, post ONE problem, tell us exactly what you don't understand, and post your work to the point that you don't know what to do next.


Thank you

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