Chemical reactions

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454 g of NH4NO3 --> N2O + 2 H2O

STEP 3 Convert moles reactant --> moles product

Relate moles NH4NO3 to moles product expected.

1 mol NH4NO3 --> 2 mol H2O

Express this relation as the STOICHIOMETRIC FACTOR.

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454 g of NH4NO3 --> N2O + 2 H2O

= 11.4 mol H2O produced

STEP 3 Convert moles reactant (5.68 mol) --> moles product

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454 g of NH4NO3 --> N2O + 2 H2O

STEP 4 Convert moles product (11.4 mol) --> mass product

Called the THEORETICAL YIELD

ALWAYS FOLLOW THESE STEPS IN SOLVING STOICHIOMETRY PROBLEMS!

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General plan for stoichiometry calculations

Stoichiometric

factor

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454 g of NH4NO3 --> N2O + 2 H2O

STEP 5 How much N2O is formed?

Total mass of reactants = total mass of products

454 g NH4NO3 = _ g N2O + 204 g H2O

mass of N2O = 250. g

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454 g of NH4NO3 --> N2O + 2 H2O

STEP 6 Calculate the percent yield

If you isolated only 131 g of N2O, what is the percent yield?

This compares the theoretical (250. g) and actual (131 g) yields.

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454 g of NH4NO3 --> N2O + 2 H2O

STEP 6 Calculate the percent yield

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PROBLEM: Using 5.00 g of H2O2, what mass of O2 and of H2O can be obtained?

2 H2O2(liq) ---> 2 H2O(g) + O2(g)

Reaction is catalyzed by MnO2

Step 1: moles of H2O2

Step 2: use STOICHIOMETRIC FACTOR to calculate moles of O2

Step 3: mass of O2

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Reactions Involving a Limiting reactant

In a given reaction, there is not enough of one reagent to use up the other reagent completely.

The reagent in short supply LIMITS the quantity of product that can be formed.

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LIMITING REACTANTS

Reactants

Products

Limiting reactant = _

Excess reactant =

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Limiting reactants

Demo of limiting reactants on Screen 4.7

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Rxn 1: Balloon inflates fully, some Zn left

* More than enough Zn to use up the 0.100 mol HCl