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**Stoichiometry - House Method**

1. Start with a balanced chemical equation. Na2CO3 + Ca(OH) NaOH + CaCO3

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**Stoichiometry - House Method**

2. Place the given information above the proper compounds in the equation. 120 g X g Na2CO3 + Ca(OH) NaOH + CaCO3

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**Stoichiometry - House Method**

3. Draw a simple house around each compound used in the problem. Add moles to the downstairs of each house. 120 g X g Na2CO3 + Ca(OH) NaOH + CaCO3 mole mole

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**Stoichiometry - House Method**

4. Draw arrows to show the path of the conversion from beginning to end. 120 g X g Na2CO3 + Ca(OH) NaOH + CaCO3 mole mole

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**Stoichiometry - House Method**

5. Set up your factor label conversions in the direction of the arrows. 120 g X g Na2CO3 + Ca(OH) NaOH + CaCO3 mole mole 120 g Na2CO3 X 1 mol Na2CO3 X mol NaOH X g NaOH = 106.0 g Na2CO mol Na2CO mol NaOH Notice that there are no numbers in front of the mol to mol conversion…YET !

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**Stoichiometry - House Method**

6. The numbers in front of the mol to mol conversion are the addresses (coefficients) of the compounds. 120 g X g 1 Na2CO3 + Ca(OH) NaOH + CaCO3 1 mole mole 2 1 120 g Na2CO3 X 1 mol Na2CO3 X mol NaOH X g NaOH = 106.0 g Na2CO mol Na2CO mol NaOH

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**Stoichiometry - House Method**

6. Solve the math. 120 g X g Na2CO3 + Ca(OH) NaOH + CaCO3 mole mole 2 1 120 g Na2CO3 X 1 mol Na2CO3 X mol NaOH X g NaOH = 106.0 g Na2CO mol Na2CO mol NaOH 90.6 g NaOH

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N2 + 3 H2 2 NH3 Calculate the number of grams of NH3 produced by the reaction with 5.40 grams of H2 with excess nitrogen.

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**Calculate the number of grams of N2 needed to produce 7.4 grams of NH3.**

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Stoichiometry Chemistry S.Fleck 2015-2016. Objectives 9.1 – The Arithmetic of Equations Interpret balanced chemical equations in terms of interacting.

Stoichiometry Chemistry S.Fleck 2015-2016. Objectives 9.1 – The Arithmetic of Equations Interpret balanced chemical equations in terms of interacting.

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