Transcript Stochiometry - House Method

Stoichiometry - House Method
1. Start with a balanced chemical equation.
Na2CO3 + Ca(OH)2
2 NaOH + CaCO3
Stoichiometry - House Method
2. Place the given information above the proper compounds in the equation.
120 g
Na2CO3 + Ca(OH)2
Xg
2 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
Xg
Na2CO3 + Ca(OH)2
mole
2 NaOH + CaCO3
mole
Stoichiometry - House Method
4. Draw arrows to show the path of the conversion from beginning to end.
120 g
Na2CO3 + Ca(OH)2
mole
Xg
2 NaOH + CaCO3
mole
Stoichiometry - House Method
5. Set up your factor label conversions in the direction of the arrows.
120 g
Na2CO3 + Ca(OH)2
mole
120 g Na2CO3 X 1 mol Na2CO3 X
106.0 g Na2CO3
Xg
2 NaOH + CaCO3
mole
mol NaOH X 40.0 g NaOH =
mol Na2CO3 1 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
Xg
1
Na2CO3 + Ca(OH)2
mole
12 NaOH
+ CaCO3
mole
120 g Na2CO3 X 1 mol Na2CO3 X 2 mol NaOH X 40.0 g NaOH =
106.0 g Na2CO3 1 mol Na2CO3 1 mol NaOH
Stoichiometry - House Method
6. Solve the math.
120 g
Na2CO3 + Ca(OH)2
mole
Xg
2 NaOH + CaCO3
mole
120 g Na2CO3 X 1 mol Na2CO3 X 2 mol NaOH X 40.0 g NaOH =
106.0 g Na2CO3 1 mol Na2CO3 1 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.