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Stoichiometry w/ Molar Mass
1. Start with a balanced chemical equation.
Na2CO3 + Ca(OH)2 2 NaOH + CaCO3
2. Place the given information above the proper compounds in the equation.
Na2CO3 + Ca(OH)2 2 NaOH + CaCO3 120 g X g
3. Draw a simple box around each compound used in the problem. Add moles to the downstairs of each house.
Na2CO3 + Ca(OH)2 2 NaOH + CaCO3 120 g X g
mole mole
4. Draw arrows to show the path of the conversion from beginning to end.
Na2CO3 + Ca(OH)2 2 NaOH + CaCO3 120 g X g
mole mole
5. Set up your factor label conversions in the direction of the arrows.
Na2CO3 + Ca(OH)2 2 NaOH + CaCO3 120 g X g
mole mole
120 g Na2CO3 X 1 mol Na2CO3 X mol NaOH X 40.0 g NaOH = 106.0 g Na2CO3 mol Na2CO3 1 mol NaOH
Notice that there are no numbers in front of the mol to mol conversion…YET !
6. The numbers in front of the mol to mol conversion are the addresses (coefficients) of the compounds.
Na2CO3 + Ca(OH)2 2 NaOH + CaCO3 120 g X g
mole mole
1 1
120 g Na2CO3 X 1 mol Na2CO3 X mol NaOH X 40.0 g NaOH = 106.0 g Na2CO3 mol Na2CO3 1 mol NaOH
21
6. Solve the math.
Na2CO3 + Ca(OH)2 2 NaOH + CaCO3 120 g X g
mole mole
120 g Na2CO3 X 1 mol Na2CO3 X mol NaOH X 40.0 g NaOH = 106.0 g Na2CO3 mol Na2CO3 1 mol NaOH
21
91 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.
