A balanced equation shows what should happen in a chemical reaction, but it does not always happen due to specific conditions such as pressure, temperature, and/or concentration.
In some reactions, one reactant will be left over, it is called the excess quantity. The second reactant is used up completely, and it's the limiting reactant(lower value).
Ex/ What mass of OCl2(g) is produced when 98.0 g of O2(g) and 134.0 g of Cl2(g) are reacted?
First we need a balanced equation for this reaction.
O2 + 2Cl2 -> 2OCl2
We take the two values of O2 and Cl2, and convert them both into grams of OCl2
98.0 g O2 x 1 mole O2 x 2 moles OCl2 x 87 g OCl2 = 533g OCl2
32 g O2 1 mole O2 1 mole OCl2
134.0g Cl2 x 1 mole Cl2 x 2 moles OCl2 x 87g OCl2 = 164g OCl2
71 g O2 2 moles Cl2 1 mole OCl2
Now we know that 164g is the limiting reactant, and that is the answer.
If we are asked how much of which reactant will be left over, we take the limiting reactant's value and convert it to the excess quantity.
134.0g Cl2 x 1 mole Cl2 x 1 mole O2 x 32 g O2 = 30.2 g O2
71 g Cl2 2 moles Cl2 1 mole O2
Finally, we take the original value and subtract it by the new value and get the amount that's in excess.
98.0 g O2 - 30.2 g O2 = 67.8 g O2 in excess
This video gives a different example and offers a good explanation as well.
(continued)
Formula: Percent yield = grams of actual product recovered x 100 %
grams of product expected from stoichiometry
Example: 2 C3H8 + 7 O2 --> 6 CO + 8 H2O
If 100.0 g of C3H8 is reacted with excess O2 and 50.0 g of CO was produced, what is the percent
yield ?
Step 1: convert C3H8 into CO
100.0 g C3H8 x 2 mol C3H8 x 6 mol CO x 18 g = 40.9 g CO
44 g C3H8 2 mol C3H8 6 mol CO
Step 2 : Use the formula.
Percent yield = 50.0 g CO recovered x 100% = 122 %
40.9 g CO obtained
Sometimes not all reactant is available to react. We calculate how much there are by using percent purity and here's the formula for it. mass of pure substance x 100 %
mass of impure sample
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