# Stoichiometry - House Method

## Presentation on theme: "Stoichiometry - House Method"— Presentation transcript:

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

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

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

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

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 !

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

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

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.

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