Presentation is loading. Please wait.

# Stoichiometric Relationships In Chemical Reactions

## Presentation on theme: "Stoichiometric Relationships In Chemical Reactions"— Presentation transcript:

Stoichiometric Relationships In Chemical Reactions
Unit 8 Stoichiometric Relationships In Chemical Reactions Mole Ratios Stoichiometry applied to gases Fractional Components of Chemical Samples Chemical Reactions Give Off or Take In Energy Limiting And Excess Reagent Problems Works best when seen as a slide show. Click the ‘Slide Show’ button on the lower right.

Mole Ratios The coefficients of a chemical equation represent a ratio of moles. They are used to balance the equation thereby reflecting the Law of Conservation of Matter. This law states that the total number of atoms during a chemical reaction does not change because matter cannot be created nor destroyed. atoms are rearranged.

2 moles of H2 react with 1 mole of O2 to get 2 moles of H2O.
Mole Ratios For the equation 2 H O2 ———> 2 H2O 2 moles of H2 react with 1 mole of O2 to get 2 moles of H2O. Furthermore, the Law of Conservation of Matter shows that hydrogen atoms yields oxygen atoms 2 water molecules [6 atoms]

Q The salt potassium chlorate undergoes decomposition when heat is applied to form the salt potassium chloride and the gas oxygen. Write the balanced equation. What is the ratio of moles: potassium chlorate to potassium chloride to oxygen? 2 3 KClO3  KCl O2 2 3

Information Given by the Chemical Equation
The coefficients in the balanced chemical equation shows the molecules and mole ratio of the reactants and products Since moles can be converted to masses, we can determine the mass ratio of the reactants and products as well

Stoichiometry Rules! Write a chemical equation. Balance the equation!
Determine start and destination for dimensional analysis. Convert start details into moles. Multiply by the appropriate mole ratio. Convert to correct units for destination.

Information Given by the Chemical Equation
2 CO + O2 —> 2 CO2 2 moles CO + 1mole O2 = 2 moles CO2 Since 1 mole of CO = g, 1 mole O2 = g, and 1 mole CO2 = g 2(28.01) g CO + 1(32.00) g O2 = 2(44.01) g CO2

Example #1 Write the balanced equation
Determine the Number of Moles of Carbon Monoxide required to react with 3.2 moles Oxygen, and determine the moles of Carbon Dioxide produced Write the balanced equation 2 CO + O2 —> 2 CO2 Use the coefficients to find the mole relationship 2 moles CO = 1 mol O2 = 2 moles CO2

Example #1 Use dimensional analysis
Determine the Number of Moles of Carbon Monoxide required to react with 3.2 moles Oxygen, and determine the moles of Carbon Dioxide produced Use dimensional analysis

Example #2 Write the balanced equation 2 CO + O2 —> 2 CO2
Determine the Number of grams of Carbon Monoxide required to react with 48.0 g Oxygen, and determine the mass of Carbon Dioxide produced Write the balanced equation 2 CO + O2 —> 2 CO2 Use the coefficients to find the mole relationship 2 moles CO : 1 mol O2 : 2 moles CO2 Determine the Molar Mass of each 1 mol CO = g 1 mol O2 = g 1 mol CO2 = g

Example #2 Determine the Number of grams of Carbon Monoxide required to react with 48.0 g Oxygen, and determine the mass of Carbon Dioxide produced Use the molar mass of the given quantity to convert it to moles Use the mole relationship to convert the moles of the given quantity to the moles of the desired quantity

Example #2 Determine the Number of grams of Carbon Monoxide required to react with 48.0 g Oxygen, and determine the mass of Carbon Dioxide produced Use the molar mass of the desired quantity to convert the moles to mass

KClO3  KCl O2 2 3 If mole of potassium chlorate decomposed, how many moles of oxygen form? If 2.04 grams of potassium chloride were produced, how many grams of potassium chlorate must have decomposed? 1.17 = mol O2 = g KClO3 3.35

Limiting And Excess Reagent Problems

One packaged meal for the USM service day:
Determine the number of packaged meals based on the given supply: 8 meals 2 Excess oranges 3 Excess sandwiches cwx.prenhall.com

In the chemical reaction
Mg(s) HCl(aq) ———> MgCl2(aq) H2(g) performed in many high school laboratories, more HCl is usually used than is needed. This is done to assure that all the magnesium reacts. The HCl, therefore, is in excess and some of it is left over when the reaction stops. No magnesium will be present once the reaction is over.

For another chemical reaction
C + O2  CO2 Initial supply of reactant Product

1. In many chemistry problems you are told which reactant is in excess and solve based on the specified quantities of the other reactant. 2. In some instances, however, you will be given quantities of both reactants. One is probably in excess, but you won’t be told which one. 3. The following problem shows one way to determine the reactant that is in excess

Q. 5. 00 g of calcium reacts with 5
Q 5.00 g of calcium reacts with 5.00 g of sulfur in a composition reaction. Do a calculation to find the a) Excess reactant. Ca(s) + S(s)  CaS(s) Higher. This means that S is in excess. In other words, after the reaction there will be CaS, unreacted S, and no Ca.

Q. 5. 00 g of calcium reacts with 5
Q 5.00 g of calcium reacts with 5.00 g of sulfur in a composition reaction. Do a calculation to find the b) Grams of product formed. Ca(s) + S(s)  CaS(s) Start with the limiting reagent since all of it will be used up. Look what happens if you start with the excess reactant:

10.0 g of original reaction mixture –
c) Calculate the grams of excess reactant unreacted once the reaction is completed. 10.0 g of original reaction mixture – 9.025 g of CaS formed = 1.0 g S left over

Determine which reactant is in excess. (ans. P)
More Practice 4.00 g of calcium is mixed with 4.00 g of phosphorus and ignited to get solid calcium phosphide, Ca3P2. Determine which reactant is in excess. (ans. P) How many grams of product were formed? (ans g) How many grams of the excess reactant were left over once the reaction was completed? (8 grams reactants – 6.05 grams calcium phosphide = 1.95 grams phosphorus left)

Similar presentations

Ads by Google