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

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

3. 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

4. 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

5. 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 !

6. 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

7. 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

8. 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.

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