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# Stoichiometry

By: Amber McMillan and Elliot Shirk
by

## Amber McMillan

on 28 September 2012

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#### Transcript of Stoichiometry

Limiting Reactant
and
Excess Reactant Limiting Reactant: The reactant that is completely used up first when creating the product(s). Definitions: 2Al + 3Cl2 ---> 2AlCl3
How many grams of aluminum chloride could be produced from 34.0 g of aluminum and 39.0 g of chlorine gas? Finding the Limiting Reactant Step 2: Now the second reactant, in grams, must be taken to grams of product it can produce. 2Al + 3Cl2 ---> 2AlCl3 Excess Reactant: This is the reactant that is not completely used up in a chemical reaction when creating the product(s). Step 1: 34.0 g Al 1 mol Al 26.98 g Al 2 mol Al 2 mol AlCl3 1 mol AlCl3 133.33 g AlCl3 168 g AlCl3 First, you take the grams of the first reactant and find out how many grams of the product it can create. 39.0 g Cl2 1 mol Cl2 68.90 g Cl2 3 mol Cl2 2 mol AlCl3 1 mol AlCl3 133.33 g AlCl3 50.3 g AlCl3 * Since the amount of AlCl3 calculated using Cl2 is less than the amount calculated from Al, Cl2 is the Limiting Reactant. * Now that we know which reactant is the limiting reactant, we also know that the remaining reactant (in this case, Al) is the excess reactant, because it is not completely used up when the product is formed. Theoretical Yield Theoretical Yield: This is the amount of product that is supposed to be formed, when calculated from the limiting reactant. Definition: 2Al + 3Cl2 ---> 2AlCl3 39.0 g Cl2 1 mol Cl2 68.90 g Cl2 3 mol Cl2 2 mol AlCl3 1 mol AlCl3 133.33 g AlCl3 50.3 g AlCl3 * We have already figured out previously what the limiting reactant is, which is Cl2. Since we know this we also know what the theoretical yield is. The amount of 50.3g AlCl3 is the theoretical yield because it was calculated from the limiting reactant, Cl2. The Remaining Excess 2Al + 3Cl2 ---> 2AlCl3 39.0 g Cl2 1 mol Cl2 68.90 g Cl2 3 mol Cl2 2 mol AlCl3 1 mol AlCl3 133.33 g AlCl3 50.3 g AlCl3 34.0 g Al 1 mol Al 26.98 g Al 2 mol Al 2 mol AlCl3 1 mol AlCl3 133.33 g AlCl3 168 g AlCl3 * We were able to figure out what the limiting and excess reactants were in the previous problem. The excess reactant is Al. Now we must figure out how much excess of Al there is. Step 1: To find the remaining excess of Al, We must first go from grams of the limiting reactant used, Cl2, to grams of Al. 2Al + 3Cl2 ---> 2AlCl3 39.0 g Cl2 1 mol Cl2 68.90 g Cl2 3 mol Cl2 2 mol Al 1 mol Al 26.98 g Al 10.2 g Al This is the amount of the excess used in the reaction. Step 2: Now we must subtract the amount used from the amount of excess given. 34.2g Al - 10.2g Al = 23.8 g Al This amount is total excess Al remaining after the reaction is taken place. Percent Yield Definition:
Percent Yield is the percentage of the theoretical yield of product that is actually produced. This allows you to see how accurate the experiment was.
Actual Yield Theoretical Yield 100 = % Yield Formula: This is the formula used to calculate percent yield. *For this example, we will use the theoretical yield calculated earlier and say that the actual yield of AlCl3 is 40.6 grams.
Actual Yield Theoretical Yield 100 % Yield =
40.6g AlCl3 50.3g AlCl3 100 80.7 % = This signifies that about 80 % of the theoretical yield of AlCl3 was actually produced. Stoichiometry
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