1. Test Review Answers

2. #1 Total Mass here is talking about molar mass.
Use the periodic table- values to 2 decimal places
Fe2O3
Iron Fe *2 =111.7
Oxygen rounds to O3= 16*3 =48
Total = = = 160g

3. #2 Percent composition = 𝑝𝑎𝑟𝑡 𝑤ℎ𝑜𝑙𝑒 ∗100
H2SO4 Percent composition of sulfur
Part= mass of sulfur 32.07*1
Whole= molar mass 1.01*2(hydrogen) ( sulfur) *4 (oxygen
Percent Composition = ∗100=32.7%

4. #3 Balanced Equation 2H2S + 3O2  2H2O + 2SO2
Sum- add them up = 9

5. #4 Use mole road map Starting at mole going to mass.
2.5 𝑚𝑜𝑙 𝐶2𝐻5𝑂𝐻 1 x 𝑔 𝐶2𝐻5𝑂𝐻 1 𝑚𝑜𝑙 𝐶2𝐻5𝑂𝐻 = 115 g C2H5OH
Molar mass, add up 2 C, 6 H, 1 O

6. #5 Which quantity is equivalent to 146 g NaCl
Use mole map- start at mass going to moles
146 𝑔 𝑁𝑎𝐶𝑙 1 x 1 𝑚𝑜𝑙 𝑁𝑎𝐶𝑙 𝑔 𝑁𝑎𝐶𝑙 =2.50 𝑚𝑜𝑙 𝑁𝑎𝐶𝑙
Molar mass, add 1 Na and 1 Cl

7. #6
What is the empirical formula of the compound whose molecular formula is P4O10
Empirical Formula is most simplified, P4O10 both numbers 4 and 10 can divide by 2, get P2O5

8. #7
Calcium reacts with sodium nitride. If 12 grams of calcium is reacted how many grams of calcium nitride is produced?
Balanced Equation
3Ca + 2NaN3  2Na + 3Ca3N2
12 𝑔 𝐶𝑎 1 𝑥 1 𝑚𝑜𝑙 𝐶𝑎 𝑔 𝐶𝑎 𝑥 3 𝑚𝑜𝑙 𝐶𝑎3𝑁2 3 𝑚𝑜𝑙 𝐶𝑎 𝑥 𝑔 𝐶𝑎3𝑁2 1 𝑚𝑜𝑙 𝐶𝑎𝑁
12x 3x /40.08/3 = 44.4 g Ca3N2

9. #8- no stoichiometry with liters on test
Methane (CH4) reacts with oxygen gas. How many liters of oxygen will be needed to react with 12 L of methane?
CH4 + 2O2  CO2 + 2H2O
12 𝐿 𝐶𝐻4 1 𝑥 1 𝑚𝑜𝑙 𝐶𝐻 𝐿 𝐶𝐻4 𝑥 2 𝑚𝑜𝑙 𝑂2 1 𝑚𝑜𝑙 𝐶𝐻4 𝑥 22.4 𝐿 𝑂2 1 𝑚𝑜𝑙 𝑂2
= 24 L O2

10. #9 Limiting Reactant: Gold (II) sulfide
Sodium Carbonate reacts with Gold (II) sulfide. If you have 3.5 grams of sodium carbonate and 4.9 grams of Gold(II) sulfide, which reactant is the limiting reactant? How much sodium sulfide will actually be produced?
Equation: Na2CO3 + AuS  Na2S + AuCO3
Known: 3.5 g Na2CO3
3.5 𝑔 𝑁𝑎2𝐶𝑂3 1 𝑥 1 𝑚𝑜𝑙 𝑁𝑎2𝐶𝑂 𝑔 𝑁𝑎2𝐶𝑂3 𝑥 1 𝑚𝑜𝑙 𝑁𝑎2𝑆 1 𝑚𝑜𝑙 𝑁𝑎2𝐶𝑂3 𝑥 𝑔 𝑁𝑎2𝑆 1 𝑚𝑜𝑙 𝑁𝑎2𝑆
= 2.58 g Na2CO3
4.9 𝑔 𝐴𝑢𝑆 1 𝑥 1 𝑚𝑜𝑙 𝑁𝑎2𝐶𝑂 𝑔 𝐴𝑢𝑆 𝑥 1 𝑚𝑜𝑙 𝑁𝑎2𝑆 1 𝑚𝑜𝑙 𝐴𝑢𝑆 𝑥 𝑔 𝑁𝑎2𝑆 1 𝑚𝑜𝑙 𝑁𝑎2𝑆
= 1.67 g Na2S
Limiting Reactant: Gold (II) sulfide
Actual amount of product 1.67 g Na2S

11. #10 Limiting Reactant: O2 Actual product made: 2.71 g H2O
Identify the limiting reactant when 2.41 g of O2 reacts with 3.67 g of H2 to produce water.
___ O2 + _2__ H2  __2_ H2O
Known: 2.41 g O2
2.41 𝑔 𝑂2 1 𝑥 1 𝑚𝑜𝑙 𝑂2 32 𝑔 𝑂2 𝑥 2 𝑚𝑜𝑙 𝐻2𝑂 1 𝑚𝑜𝑙 𝑂2 𝑥 𝑔 𝐻2𝑂 1 𝑚𝑜𝑙 𝐻2𝑂 =2.71 𝑔 𝐻2𝑂
Known: 3.67 g H2
3.67 𝑔 𝐻2 1 𝑥 1 𝑚𝑜𝑙 𝐻 𝑔 𝐻2 𝑥 2 𝑚𝑜𝑙 𝐻2𝑂 2 𝑚𝑜𝑙 𝐻2 𝑥 𝑔 𝐻2𝑂 1 𝑚𝑜𝑙 𝐻20 =32.7 𝑔 H2O
Limiting Reactant: O2
Actual product made: 2.71 g H2O

12. #11
What mass of SO2 is produced from the reaction between 24.6 g of S8 and 8.65 g of O2?
___ S8 + __8_ O2  __8_ SO2
Known: 24.6 g S8
24.6 𝑔 𝑆8 1 𝑥 1 𝑚𝑜𝑙 𝑆 𝑔 𝑆8 𝑥 8 𝑚𝑜𝑙 𝑆𝑂2 1 𝑚𝑜𝑙 𝑆8 𝑥 𝑔 𝑆𝑂2 1 𝑚𝑜𝑙 𝑆𝑂2 = 49.1 g SO2
Known: 8.65 g O2
8.65 𝑔 𝑂2 1 𝑥 1 𝑚𝑜𝑙 𝑂2 32 𝑔 𝑂2 𝑥 8 𝑚𝑜𝑙 𝑂2 8 𝑚𝑜𝑙 𝑆𝑂2 𝑥 𝑔 𝑆𝑂2 1 𝑚𝑜𝑙 𝑆𝑂2 = 17.3 g SO2
Limiting Reactant: O2
Actual product amount: 17.3 g SO2

13. #12
Determine the percent yield for the reaction between 3.74 g of Na and excess O2 if 5.34 g of Na2O2 is recovered.
__2_ Na + ___ O2  ___ Na2O2
Actual Yield: 5.34 g
Theoretical Yield: calculate using stoichiometry
Known: 3.74 g Na Want: g Na2O2=
3.74 𝑔 𝑁𝑎 1 𝑥 1 𝑚𝑜𝑙 𝑁𝑎 23.00𝑔 𝑁𝑎 𝑥 1 𝑚𝑜𝑙 𝑁𝑎2𝑂2 2 𝑚𝑜𝑙 𝑁𝑎 𝑥 𝑔 𝑁𝑎2𝑂2 1 𝑚𝑜𝑙 𝑁𝑎2𝑂2 =6.34 𝑔 𝑁𝑎2𝑂2
Percent Yield= 𝑎𝑐𝑡𝑢𝑎𝑙 𝑦𝑖𝑒𝑙𝑑 𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑦𝑖𝑒𝑙𝑑 𝑥100
Percent Yield = 5.34g /6.34 g x100 = 84.2%

14. #13- nothing this hard on test
Determine the percent yield for the reaction between 82.4 g of Rb and g of O2 if 39.7 g of RbO2 is produced.
_1__ Rb + _1__ O2  __1_ RbO2
Actual Yield: 39.7 g RbO2
Theoretical Yield: Do a limiting reactant problem
82.4 𝑔 𝑅𝑏 1 𝑥 1 𝑚𝑜𝑙 𝑅𝑏 𝑔 𝑅𝑏 𝑥 1 𝑚𝑜𝑙 𝑅𝑏𝑂2 1 𝑚𝑜𝑙 𝑅𝑏 𝑥 𝑔 𝑅𝑏𝑂2 1 𝑚𝑜𝑙 𝑅𝑏𝑂2 =112 𝑔 𝑅𝑏𝑂2
13.49 𝑔 𝑂2 1 𝑥 1 𝑚𝑜𝑙 𝑂2 32 𝑔 𝑂2 𝑥 1 𝑚𝑜𝑙 𝑅𝑏𝑂2 1 𝑚𝑜𝑙 𝑂2 𝑥 𝑔 𝑅𝑏𝑂2 1 𝑚𝑜𝑙 𝑅𝑏𝑂2 =49.1 𝑔 𝑅𝑏𝑂2
Theoretical Yield = 49.1 g RbO2
Percent Yield= 39.7 g / 49.1 g x100 = 80.9%