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MoleConversions
VocabularyMultiple-StepConversions
OtherCalculations
This is the number of grams in 1.25 moles of copper.
A. 0.0197 g Cu B. 28.0 g Cu C. 79.4 g Cu D. 7.53 x 1023 g Cu
?1 $200
This is the number of moles in 2.67 x 1025 atoms of zinc.
A. 44 moles ZnB. 44.4 moles ZnC. 2900 moles ZnD. 1.61 x 1049 moles Zn
?1 $300
This is the volume of 8.53 mol of carbon dioxide at STP.
A. 0.381 L CO2
B. 191 L CO2
C. 4.34 L CO2
D. 8410 L CO2
?1 $400
This is the number of moles of iron needed to react completely with 8.20 moles of oxygen in the reaction below:4 Fe + 3 O2 → 2 Fe2O3
A. 1.82 x 10-23 moles Fe B. 6.15 moles FeC. 10.9 moles FeD. 610. moles Fe
?1 $500
?2 $100
This is the name for the number of particles in one mole (6.02 x 1023).
A. Avogadro’s numberB. Gay Lussac’s numberC. Charles’ number
This is the simplest ratio that represents the lowest whole number ratio of elements in a compound.
A. Molecular FormulaB. Limiting ReagentC. Theoretical YieldD. Empirical Formula
?2 $200
?2 $300
This is the sum of the atomic masses of all the atoms in a compound.
A. Empirical FormulaB. Molar MassC. Avogadro’s numberD. Percent Yield
?2 $400This is the maximum amount of product that can be obtained from a given amount of reactants in a chemical reaction.
A. Percent YieldB. Actual YieldC. Theoretical YieldD. Percent Error
?2 $500This is the branch of chemistry and chemical engineering that deals with the quantities of substances that enter into, and are produced by, chemical reactions.
A. Analytical chemistryB. Conservation chemistryC. StoichiometryD. Quantum mechanics
This is the volume of O2 formed when 3.50 moles of water decomposes to hydrogen and oxygen in the following equation:2 H2O → 2 H2 + O2
A. 39.2 L O2B. 39.2 L H2 C. 78.4 L O2 D. 78.4 L H2
?3 $100
This is the number of grams of potassium chloride formed when 27.8 g of potassium chlorate reacts in the following equation: 2 KClO3 → 2 KCl + 3 O2
A. 11.3 g KCl B. 16.9 g KCl C. 25.4 g KCl D. 45.7 g KCl
?3 $200
?3 $300
This is the number of moles of H2O formed when 33.0 grams of H2SO4 reacts with KOH.H2SO4 + 2 KOH → K2SO4 + 2 H2O
A. 0.168 moles H2O B. 0.673 moles H2OC. 3.03 moles H2O D. 12.1 moles H2O
This is the theoretical yield of sodium chloride when 19.4 g of sodium react with chlorine in the following equation:2 Na + Cl2 → 2 NaCl
A. 49.3 g B. 7.63 gC. 24.7 gD. 98.7 g
?3 $400
This is the reason why the number of grams of oxygen that reacted can be calculated without dimensional analysis in the following lab data set: 4 Fe + 3 O2 → 2 Fe2O3 Lab DataMass of iron reacted: 67.4 gMass of iron (III) oxide formed: 113.9 g
A. Law of Conservation of MassB. Percent yield is 100%C. Percent error is 0%D. B & C
?3 $500
A3 $500
What is 46.5 g of O2 reacted because of the Law of Conservation of Mass which states that the amount of mass
must be conserved in a chemical reaction. The sum of the masses of
the reactants must equal the mass of the products.
This is the formula in the list below which is not an empirical formula
A. K2SB. C4H10
C. Li2SO4
D. H3PO4
?4 $200
This is the percent yield obtained if the theoretical yield of magnesium oxide is 17.2 g from a mass-mass calculation, but a chemistry student produced 13.8 g.
A. 16.4 %B. 24.6 %C. 80.2 %D. 125 %
?4 $300
Find the empirical formula of a compound that is 68.5% carbon, 8.6% hydrogen, and 22.8% oxygen.
This is the subscript for carbon.
?4 $500
What is 4 - C4H6O?
C – 68.5 / 12.0 = 5.7 / 1.4 = 4 (4.07)H – 8.6 / 1.0 = 8.6 / 1.4 = 6 (6.14)O – 22.8 / 16.0 =1.4 / 1.4 = 1
A4 $500
This is the mass of reactant used in the following reaction: 2Mg + O2 → 2MgOData Table
A. 2.26 gB. 5.71 g
?5 $100
Constant mass of evaporating dish 27.88 g
Mass of evaporating dish and Mg 30.14 g
Mass of evaporating dish and MgO (first heating)
33.65 g
Mass of evaporating dish and MgO (second heating)
33.59 g
Constant mass of evaporating dish and MgO
33.59 g
This is the experimental yield of MgO from the following reaction: 2Mg + O2 → 2MgOData Table
A. 2.26 gB. 5.71 g
?5 $200
Constant mass of evaporating dish 27.88 g
Mass of evaporating dish and Mg 30.14 g
Mass of evaporating dish and MgO (first heating)
33.65 g
Mass of evaporating dish and MgO (second heating)
33.59 g
Constant mass of evaporating dish and MgO
33.59 g
This is a reason why the theoretical ratio of a hydrate is often lower than the experimental ratio.
A. Humidity in the environmentB. Insufficient heatingC. Calculator errorD. A & B
?5 $300
These are reasons why experimental yield is often less than theoretical yield.
A. The reaction does not go to completionB. Product may be lost due to poor technique (spilling, transfer, etc.)C. Reactants may not be pureD. Side reactions may occurE. All of the above
?5 $400
Calculate the percent yield of the equation below using the information in the data table.2Mg + O2 → 2MgO
A. 91.1 %B. 110. %
?5 $500
Constant mass of evaporating dish 31.95 g
Mass of evaporating dish and Mg 35.12 g
Mass of evaporating dish and MgO (first heating)
36.77 g
Mass of evaporating dish and MgO (second heating)
36.74 g
Constant mass of evaporating dish and MgO 36.74 g
Mass of reactant = 35.12 – 31.95 = 3.17 g Mg
Theoretical Yield = 5.26 g MgO
Actual Yield = 36.74 – 31.95 = 4.79 g MgO
Percent Yield = 4.79/5.26 * 100 = 91.1%
Constant mass of evaporating dish 31.95 g
Mass of evaporating dish and Mg 35.12 g
Mass of evaporating dish and MgO (first heating)
36.77 g
Mass of evaporating dish and MgO (second heating)
36.74 g
Constant mass of evaporating dish and MgO 36.74 g