Test Review Answers

#1 Total Mass here is talking about molar mass. Use the periodic table- values to 2 decimal places Fe2O3 Iron- 55.85 Fe2 55.85*2 =111.7 Oxygen 15.998 rounds to 16.00 O3= 16*3 =48 Total = 111.7+48 = 159.7 = 160g

#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) +32.07( sulfur) +16.00*4 (oxygen Percent Composition = 32.07 98.09 ∗100=32.7%

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

#4 Use mole road map Starting at mole going to mass. 2.5 𝑚𝑜𝑙 𝐶2𝐻5𝑂𝐻 1 x 46.08 𝑔 𝐶2𝐻5𝑂𝐻 1 𝑚𝑜𝑙 𝐶2𝐻5𝑂𝐻 = 115 g C2H5OH Molar mass, add up 2 C, 6 H, 1 O

#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 58.45 𝑔 𝑁𝑎𝐶𝑙 =2.50 𝑚𝑜𝑙 𝑁𝑎𝐶𝑙 Molar mass, add 1 Na and 1 Cl

#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

#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 40.08 𝑔 𝐶𝑎 𝑥 3 𝑚𝑜𝑙 𝐶𝑎3𝑁2 3 𝑚𝑜𝑙 𝐶𝑎 𝑥 148.28 𝑔 𝐶𝑎3𝑁2 1 𝑚𝑜𝑙 𝐶𝑎𝑁 12x 3x 148.28 /40.08/3 = 44.4 g Ca3N2

#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 𝐶𝐻4 22.4 𝐿 𝐶𝐻4 𝑥 2 𝑚𝑜𝑙 𝑂2 1 𝑚𝑜𝑙 𝐶𝐻4 𝑥 22.4 𝐿 𝑂2 1 𝑚𝑜𝑙 𝑂2 = 24 L O2

#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 𝑁𝑎2𝐶𝑂3 106.01 𝑔 𝑁𝑎2𝐶𝑂3 𝑥 1 𝑚𝑜𝑙 𝑁𝑎2𝑆 1 𝑚𝑜𝑙 𝑁𝑎2𝐶𝑂3 𝑥 78.07 𝑔 𝑁𝑎2𝑆 1 𝑚𝑜𝑙 𝑁𝑎2𝑆 = 2.58 g Na2CO3 4.9 𝑔 𝐴𝑢𝑆 1 𝑥 1 𝑚𝑜𝑙 𝑁𝑎2𝐶𝑂3 229.04 𝑔 𝐴𝑢𝑆 𝑥 1 𝑚𝑜𝑙 𝑁𝑎2𝑆 1 𝑚𝑜𝑙 𝐴𝑢𝑆 𝑥 78.07 𝑔 𝑁𝑎2𝑆 1 𝑚𝑜𝑙 𝑁𝑎2𝑆 = 1.67 g Na2S Limiting Reactant: Gold (II) sulfide Actual amount of product 1.67 g Na2S

#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 18.02 𝑔 𝐻2𝑂 1 𝑚𝑜𝑙 𝐻2𝑂 =2.71 𝑔 𝐻2𝑂 Known: 3.67 g H2 3.67 𝑔 𝐻2 1 𝑥 1 𝑚𝑜𝑙 𝐻2 2.02 𝑔 𝐻2 𝑥 2 𝑚𝑜𝑙 𝐻2𝑂 2 𝑚𝑜𝑙 𝐻2 𝑥 18.02 𝑔 𝐻2𝑂 1 𝑚𝑜𝑙 𝐻20 =32.7 𝑔 H2O Limiting Reactant: O2 Actual product made: 2.71 g H2O

#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 𝑆8 256.56 𝑔 𝑆8 𝑥 8 𝑚𝑜𝑙 𝑆𝑂2 1 𝑚𝑜𝑙 𝑆8 𝑥 64.07 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 64.07 𝑔 𝑆𝑂2 1 𝑚𝑜𝑙 𝑆𝑂2 = 17.3 g SO2 Limiting Reactant: O2 Actual product amount: 17.3 g SO2

#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 78.00 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%

#13- nothing this hard on test Determine the percent yield for the reaction between 82.4 g of Rb and 13.49 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 85.47 𝑔 𝑅𝑏 𝑥 1 𝑚𝑜𝑙 𝑅𝑏𝑂2 1 𝑚𝑜𝑙 𝑅𝑏 𝑥 116.47 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 116.47 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%