Limiting+Reactants

=Limiting Reactant= The limiting reactant is the reactant that runs out first in a chemical reaction, causing the reaction to stop. It limits how much product can be made, which is why it is called the limiting reactant. It then leaves all the other reactants in the experiment which is called the Excess Reactants. http://www.elements.nb.ca/theme/ecologicalfootprint/ray/cheeseburger.jpg

The limiting reactant is comparable to making ten of cheeseburgers because you need one burger, two buns, one piece of lettuce, and one slice of cheese to make one basic cheeseburger. So if you plan on making ten cheeseburgers, and have 10 burgers, 11 pieces of cheese, 20 buns, and nine leaves of lettuce, you will run out of lettuce before you finish, so the lettuce will be the limiting reactant. And, because we have eleven pieces of cheese, the cheese will be an Excess Reactant, since we will only use 10 pieces of cheese.

Solving for Limiting Reactants: If a reaction such as 2NaOH + CO2 -> Na2CO3 + H20 were used in an experiment containing 1.70g of NaOH and 1.00g of CO2, then what would be the limiting reactant?

1. First, one would first convert everything into moles, since reactants react in a mole ratio, which is given in the chemical equation, we can't use mass to find the limiting reactant. NaOH's molar mass is [23.0g/mole Na + 16.0g/mole O +1.01g/mole H] = 40.0g/mole NaOH CO2's molar mass is [12.0g/mole + (16.0g/mole x 2)] = 44.0g/mole of CO2 2. Now that we know how many moles of each reactant we have, we can use the chemical equation to find the reactant to product mole ratio, so that we can calculate how many moles of product can be made if one of the certain reactants was not limited by the other reactants. This way, when we do find the moles of product each reactant can make, the one that can make the least product is the limiting reactant, since it will run out once it makes that many products. 2NaOH + CO2 ---> Na2CO3 + H20

One can see that the amount of product, Na2CO3, from NaOH, 0.02125 moles, is less than from CO2 , 0.0227 moles. This means that NaOH is the limiting reactant.

3. If the question asked how much product was made from this specific experiment, all one would have to do to find that would be to take the number of moles of Na2CO3 from the NaOH equation and find the weight of that via molar mass and factor label.

Na2CO3's molar mass is [(23.0g/mole Na x 2) + 12.0g/mole C + (16.0g/mole O x 3)] = 106.0g/mole Na2CO3.

4. Finding the amount of excess reactant is the same as finding the product. Use the mole to mole ratio provided by the chemical equation and find how many moles of excess reactant was used in the experiment and them convert it into mass if need be. However that is the amount, used so subtract that from the starting mass to get the mass left over. 1.00g - 0.935g = 0.065 grams of CO2 left All these calculations are theoretical yields.

=Percent Yield : 100 x (actual yield / theoretical yield)=

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The limiting reagent you found in the problem above is the theoretical yield, that is the amount you would get in a perfect world. However, because of slight errors this amount is very rarely obtained.=====