Answer:
Theoretical Yield
Percent yield
Example stoichiometry problem
How much oxygen can be prepared from 12.25 g KClO3 . (Use molar mass KClO3 = 122.5 g.)
Most stoichiometry problems can be solved using the following steps.
Step 1.
Write and balance the equation for the decomposition of KClO3 with heat (∆). 2KClO3 + ∆ → 2KCl + 3O2
Step 2.
Convert what you have (in this case g KClO3) to moles.
# moles = grams/molar mass = 12.25 g /122.5 = 0.100 mole KClO3.
Step 3.
Using the coefficients in the balanced equation, convert moles of what you have (moles KClO3) to moles of what you want (in this case moles oxygen).
0.100 mol KClO3 x (3 moles O2/2 moles KClO3) = 0.100 x (3/2) = 0.150 mole O2.
Step 4.
Convert moles from step 3 to grams.
moles x molar mass = grams
0.150 mole O2 x (32.0 g O2/mole O2) = 4.80 g O2 produced from 12.25 g KClO3. This is the theoretical yield. If the ACTUAL yield is 4.20 grams, calculate percent yield. Percent yield = (actual yield/theoretical yield) x 100 = (4.20/4.80) x 100 = 87.5% yield
NOTE: In step 1, moles can be obtained other ways; in step 4 moles can be converted to other units.
a. For solutions, M x L = moles (or mL x M = millimoles).
b. For gases, L/22.4 = moles
solubility will increase, since it is basic in nature, on acidifying solubility increases.
being basic on acidification, the solubility increases.
is alkaline in nature, it gets neutralized and there is no increase in acidity.
is a salt and it shows no effect on acidification.
is a salt but insoluble in water and shows no effect on acidification.
is strongly basic in nature, on acidifying the solubility increases.