1. Stoichiometry w/ Molar Mass 1. Start with a balanced chemical equation. Na 2 CO 3 + Ca(OH) 2 2 NaOH + CaCO 3

2. 2. Place the given information above the proper compounds in the equation. Na 2 CO 3 + Ca(OH) 2 2 NaOH + CaCO 3 120 gX g

3. 3. Draw a simple box around each compound used in the problem. Add moles to the downstairs of each house. Na 2 CO 3 + Ca(OH) 2 2 NaOH + CaCO 3 120 gX g mole

4. 4. Draw arrows to show the path of the conversion from beginning to end. Na 2 CO 3 + Ca(OH) 2 2 NaOH + CaCO 3 120 gX g mole

5. 5. Set up your factor label conversions in the direction of the arrows. Na 2 CO 3 + Ca(OH) 2 2 NaOH + CaCO 3 120 gX g mole 120 g Na 2 CO 3 X 1 mol Na 2 CO 3 X mol NaOH X 40.0 g NaOH = 106.0 g Na 2 CO 3 mol Na 2 CO 3 1 mol NaOH Notice that there are no numbers in front of the mol to mol conversion…YET !

6. 6. The numbers in front of the mol to mol conversion are the addresses (coefficients) of the compounds. Na 2 CO 3 + Ca(OH) 2 2 NaOH + CaCO 3 120 gX g mole 1 1 120 g Na 2 CO 3 X 1 mol Na 2 CO 3 X mol NaOH X 40.0 g NaOH = 106.0 g Na 2 CO 3 mol Na 2 CO 3 1 mol NaOH 2121

7. 6. Solve the math. Na 2 CO 3 + Ca(OH) 2 2 NaOH + CaCO 3 120 gX g mole 120 g Na 2 CO 3 X 1 mol Na 2 CO 3 X mol NaOH X 40.0 g NaOH = 106.0 g Na 2 CO 3 mol Na 2 CO 3 1 mol NaOH 2121 91 g NaOH

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

9. Calculate the number of grams of N 2 needed to produce 7.4 grams of NH 3.