Balancing
Classifying
The Mole/Chm formulas
Stoichiometry/% yield
Limiting reactant
100

Balance: H2 + NO →  H2O  +  N2 

2 H2 + 2 NO → 2 H2O + N2

100

Br2 + 2 NH4Cl → 2 NH4Br + Cl2

single replacement

100
What is Avogadro's number?

6.02 x 1023

100

2 FeCl2 + Cl2 → 2 FeCl3
Given: 75.18 grams FeCl2 reacted
Wanted: grams of Cl2 reacted?  

21.03 grams Cl2

100

6Na + Fe2O3 --> 3Na2O + 2Fe If you are provided 200g of sodium and 250 grams of iron(III) oxide, which substance is the limiting reactant?

sodium

200

Balance: Cu(NO3)2 + KOH →  Cu(OH)2  +  KNO3

Cu(NO3)2 + 2 KOH → Cu(OH)2 + 2 KNO3

200

2 Ca + O2 → 2 CaO

combination

200

What is molar mass and how is it found? 

The mass of one mole of a substance, add together the masses of each element from the chemical formula

200

NH4OH → NH3 + H2O
Given: 412.19 grams NH4OH reacted
Wanted: grams of H2O produced?

211.89 grams H2O

200

6Na + Fe2O3 --> 3Na2O + 2Fe If you are provided 200g of sodium and 250 grams of iron(III) oxide, determine the amount of solid iron produced.

162 g Fe

300

Balance: Al2O3 + C →  Al  +  CO2

2 Al2O3 + 3 C → 4 Al + 3 CO2

300

Cu(NO3)2 + 2 KOH → Cu(OH)2 + 2 KNO3

double replacement

300

Determine empirical formula of:  40.0% carbon, 6.7% hydrogen, and 53.3% oxygen 

CH2O

300

2 Al + 3 H2SO4 → Al2(SO4)3 + 3 H2
Given: 1.0186 moles H2SO4 reacted
Wanted: moles of Al2(SO4)3 produced?

0.33953 moles Al2(SO4)3

300

6Na + Fe2O3 --> 3Na2O + 2Fe If you are provided 200g of sodium and 250 grams of iron(III) oxide, how much of the iron(III) oxide was left over?

18 g

400

Balance: C6H10O5 + O2 →  CO2  +  H2O

C6H10O5 + 6 O2 → 6 CO2 + 5 H2O

400

4 KNO3 → 2 K2O + 2 N2 + 5 O2

decomposition

400

Empirical formula: CH2O

Formula mass 180 g/mol

C6H12O6

400

4 Si2H3 + 11 O2 → 8 SiO2 + 6 H2O
Given: 8.8 × 1022 particles SiO2 produced, 1.6 × 1023 particles of O2 reacted
Wanted: % yield?

Correct answer is 76%
100% yield would give 1.2 × 1023 particles SiO2

400

If you have 3.50 moles of hydrogen and 5.00 moles of nitrogen to produce ammonia, how much nitrogen is used up? 3H2 + N2 --> 2NH3

32.7 g

500

Balance: C5H10O5 + O2 →  CO2  +  H2O

C5H10O5 + 5 O2 → 5 CO2 + 5 H2O

500

2 C6H6 + 15 O2 → 12 CO2 + 6 H2O

combustion

500

When is mass in grams the same as mass in amu?

When there is one mole of a substance

500

F2 + 2 KCl → 2 KF + Cl2
Given: 1.6 × 1025 particles F2 reacted at 67% yield
Wanted: particles of KF produced?

Correct answer is 2.1 × 1025 particles KF
• 100% yield would give 3.2 × 1025 particles KF

500

If you have 3.50 moles of hydrogen and 5.00 moles of nitrogen to produce ammonia, how much nitrogen is left over? 3H2 + N2 --> 2NH3

107.3 g

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