Stoichiometry - House Method 1. Start with a balanced chemical equation.

Na 2 CO 3 + Ca(OH) 2 2 NaOH + CaCO 3

Stoichiometry - House Method 2. Place the given information above the proper compounds in the equation.

120 g

Na 2 CO 3 + Ca(OH) 2

X g

2 NaOH + CaCO 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

Na 2 CO 3 + Ca(OH) 2 mole 2 NaOH + CaCO mole 3

Stoichiometry - House Method 4. Draw arrows to show the path of the conversion from beginning to end.

120 g

Na 2 CO 3 + Ca(OH) 2 mole

X g

2 NaOH + CaCO 3 mole

Stoichiometry - House Method 5. Set up your factor label conversions in the direction of the arrows.

120 g

Na 2 CO 3 + Ca(OH) 2 mole

X g

2 NaOH + CaCO 3 mole 120 g Na 2 CO 3 X 1 mol Na 2 CO 3 106.0 g Na 2 CO 3 X mol NaOH X 40.0 g NaOH = mol Na 2 CO 3 1 mol NaOH Notice that there are no numbers in front of the mol to mol conversion…YET !

Stoichiometry - House Method 1 6. The numbers in front of the mol to mol conversion are the addresses (coefficients) of the compounds.

120 g X g

Na 2 CO mole 3 + Ca(OH) 2 1 2 NaOH + CaCO 3 mole 120 g Na 2 CO 3 X 1 mol Na 2 CO 3 106.0 g Na 2 CO 3 X 2 1 mol NaOH X 40.0 g NaOH = mol Na 2 CO 3 1 mol NaOH

Stoichiometry - House Method 6. Solve the math.

120 g

Na 2 CO 3 + Ca(OH) 2 mole

X g

2 NaOH + CaCO 3 mole 120 g Na 2 CO 3 X 1 mol Na 2 CO 3 106.0 g Na 2 CO 3 X 2 1 mol NaOH X 40.0 g NaOH = mol Na 2 CO 3 1 mol NaOH 90.6 g NaOH

• • N 2 + 3 H 2  2 NH 3 Calculate the number of grams of NH 3 the reaction with 5.40 grams of H 2 produced by with excess nitrogen.

• Calculate the number of grams of N 2 produce 7.4 grams of NH 3 .

needed to