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Limiting Reactants and Stoichiometry

Limiting Reactants and Stoichiometry. Honors Chemistry. Limiting Reactants. In a chemical reaction, the amount of product that can be made depends on the amount of reactant available. The reactant that determines the amount of product that can be made is called the limiting reactant .

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Limiting Reactants and Stoichiometry

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  1. Limiting Reactants and Stoichiometry Honors Chemistry

  2. Limiting Reactants • In a chemical reaction, the amount of product that can be made depends on the amount of reactant available. • The reactant that determines the amount of product that can be made is called the limiting reactant. • Example – If I have 1000 of each of the parts to make a bicycle, but only 10 wheels, how many bicycles can I make? • 5 bicycles • What is the limiting part? • THE WHEELS!!!!

  3. Basic Steps to finding the limiting reactant in a chemical equation: • Write a balanced equation. 2. Do a mass to mass problem using the starting amount for each reactant. 3. The limiting reactant is the reactant that gives you the smaller amount of product. (The smaller answer will be the correct answer!!!)

  4. Ex. How many grams of Zinc Chloride can be made when 5.0 g of Hydrochloric Acid reacts with 5.0 g of Zinc metal. 1. Write a balanced equation. 2HCl + Zn ---> ZnCl2 + H2 2. Do a mass to mass problem using the starting amount for each reactant. 5.0 g HCl 1 mole HCl 1 mole ZnCl2 134.4 g ZnCl2 = 9.4 g ZnCl2 36.5 g HCl 2 moles HCl 1 mole ZnCl2 5.0 g Zn 1 mole Zn 1 mole ZnCl2 134.4 g ZnCl2 = 10.3 g ZnCl2 65.4 g Zn 1 mole Zn 1 mole ZnCl2

  5. The limiting reactant is the reactant that gives you the smaller amount of product. Therefore HCl is the limiting reactant in this problem. • You do not have enough HCl to react away all the zinc that was added to the beaker • 5.0 g HCl 1 mole HCl 1 mole ZnCl2 134.4 g ZnCl2 = 9.4 g ZnCl2 36.5 g HCl 2 moles HCl 1 mole ZnCl2 • 5.0 g Zn 1 mole Zn 1 mole ZnCl2 134.4 g ZnCl2 = 10.3 gZnCl2 • 65.4 g Zn 1 mole Zn 1 mole ZnCl2 In this reaction 9.4 grams of ZnCl2 will form and there will be leftover zinc that did not react. The HCl will be all used up!!!

  6. To find out how much of the excess reactant is left over, start with the initial mass of the limiting reactant and do a mass to mass problem to determine how much of the excess reactant was needed 2HCl + Zn ---> ZnCl2 + H2 5.0 grams HCl 1 mole HC 1 mole Zn 65.4 g Zn = 4.48 g Zn needed 36.5 g HCl 2 moles HCl 1 mole Zn Subtract this value from the initial mass of the excess reactant. *4.48 grams is how much zinc reacted away given the amount of HCl available. We used 5.0 grams in the reaction, so . . . . . 5.0 grams – 4.48 grams = 0.52 grams Zn left over.

  7. Limiting Reactant Example Problem: • Silver Nitrate and sodium phosphate are reacted in equal amounts of 200.grams each. How many grams of silver phosphate are produced? 3AgNO3 + Na3PO4 Ag3PO4 + 3NaNO3

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