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Limiting reagents

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Equation balancing & stoichiometry lectures » limiting reagent calculations

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Limiting reagent calculations are not much more difficult than any other stoichiometric calculations, there is just one step more - comparison of amounts of reacting substances.

Once we have balanced reaction equation we know what amounts of substances will react. Let's try the following reaction of calcium hydroxide synthesis:

CaO + H2O -> Ca(OH)2

One mole of calcium oxide reacts with one mole of water to produce one mole of calcium hydroxide. What will happen if we have one mole of calcium oxide and two moles of water? Well, nothing will change - still only one mole of CaO can react with H2O, so after the reaction there will be one mole of water left. We call the reagent that reacts to the end limiting reagent.

Let's try to find out how much calcium hydroxide will be produced when 10.00 g of calcium oxide reacts with 3.00 g of water.

First of all we have to check which one is limiting reagent. How much water will react with 10.0 grams of CaO?

56.08 g:18.01 g
10.00 g:x g

56.08 g : 18.01 g = 10.0 g : x g

x = 10.00 g/56.08 g×18.01 g = 3.213 g

We have not enough water - 3.00 g only, when CaO needs 3.213 g, so water is a limiting reagent. How much calcium hydroxide will be produced from 3.00 g of water?

18.01 g:74.09 g
3.00 g:x g

18.01 g : 74.09 g = 3.00 g : x g

x = 3.00 g/18.01 g×74.09 g = 12.34 g

That's the final answer.

Using EBAS - equation balancer and stoichiometry calculator - you can check which substance is the limiting reagent just entering known amounts - the one displayed red is in excess.

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