Balancing Equations and Stoichiometry. Chemical Equations Terms: (s) = solid (l) = liquid (g) = gas  = heat (aq) = aqueous solution.

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Presentation transcript:

Balancing Equations and Stoichiometry

Chemical Equations Terms: (s) = solid (l) = liquid (g) = gas  = heat (aq) = aqueous solution

Balancing Equations * Use coefficients to balance equations! Step 1: Balance metals first. Step 2: If possible, consider poly-atomic ions as a group. If “OH” is present on one side and H 2 O is present on the other side, break up water into H and OH.

Balancing Equations Step 3: Balance other elements Step 4: Balance H’s and O’s last. Step 4: Double-check.

Sample Problem Balance the reaction: Cu + AgNO 3 Ag + Cu(NO 3 ) 2 Ca(OH) 2 + H 3 PO 4 H 2 O + Ca 3 (PO 4 ) 2

Stoichiometric Calculations Given the reaction: C 3 H 8 +5O 2 CO 2 + 4H 2 O Info:molar ratios

Problem C 3 H 8 +O 2 CO 2 + 4H 2 O If 25 grams of C 3 H 8 is used, how much O 2 is needed?

Before anything else…. I. Find the molar masses of each compound in the reaction. II. Calculate the mole of the compound given using the equation: mole = gram / molar mass

Solution 1. Balance equation. 2. Get molar ratios from balanced equation. 3. Find actual moles using given masses.

Solution (cont.) 4. Re-adjust moles. 5. Convert moles to grams if required.

Steps in Stoichiometry 1. Get the molar masses of each cpd in the equation. 2. If grams are given, convert grams to moles using the equation: mole = gram/molar mass 3. Balance the equation.

4. If only 1 mass is given (Case I), there is no limiting reagent. Re-adjust each mole using the molar ratios from the balanced equation. 5. If more than 1 mass is given, there is a LIMITING REAGENT! Base all actual moles of needed reactant and desired product on the Limiting Reagent (not on the Excess)! (Case II)

How to IDENTIFY the Limiting Reagent Divide the Actual Moles by the coefficents for that compound from the balanced equation. The smaller quotient is the Limiting Reagent. (This is for identification only. Cross out these values afterwards.) Use the ACTUAL MOLE of the LR in your calculations to re- adjust the ratios.

6. Convert all moles to grams using the equation: gram = mole x molar mass 7. Grams of Reactants must equal grams of products.

Limiting and Excess Reagents Limiting reagent = limits the amt. of product that can form Excess Reagent = reagent that is over and above what is needed

Case II Stoichiometry Has a limiting and excess reagent Case II applies when there are 2 or more given masses or moles

5. Convert moles to grams, if needed. Gram = mole x molar mass 6. Calculate % Yield and % Error, if needed.

Determining the Limiting Reagent To determine the limiting reagent, divide all calculated moles by the coefficients in the balanced reaction. The smallest value is the Limiting Reagent. Please note: Do not use these values for the rest of your calculations. This is only for the IDENTIFICATION of the Limiting Reagent!

Yields Theoretical Yield –the amount of product formed when the limiting reagent is totally consumed

Yield Actual Yield - often given as percent yield % Yield = actual yield X 100 theoretical yield