Lab Questions Mole Conversions Vocabulary Multiple-Step Conversions Other Calculations

This is the number of particles in one mole of sodium. A. 1 B C D x ?1 $100

This is the number of grams in 1.25 moles of copper. A g Cu B g Cu C g Cu D x g Cu ?1 $200

A1 $200

This is the number of moles in 2.67 x atoms of zinc. A. 44 moles Zn B moles Zn C moles Zn D x moles Zn ?1 $300

A1 $300

This is the volume of 8.53 mol of carbon dioxide at STP. A L CO 2 B. 191 L CO 2 C L CO 2 D L CO 2 ?1 $400

A1 $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 O 2 → 2 Fe 2 O 3 A x moles Fe B moles Fe C moles Fe D moles Fe ?1 $500

A1 $500

?2 $100 This is the name for the number of particles in one mole (6.02 x ). A.Avogadro’s number B.Gay Lussac’s number C.Charles’ number

This is the simplest ratio that represents the lowest whole number ratio of elements in a compound. A. Molecular Formula B. Limiting Reagent C. Theoretical Yield D. Empirical Formula ?2 $200

?2 $300 This is the sum of the atomic masses of all the atoms in a compound. A. Empirical Formula B. Molar Mass C. Avogadro’s number D. Percent Yield

?2 $400 This is the maximum amount of product that can be obtained from a given amount of reactants in a chemical reaction. A.Percent Yield B.Actual Yield C.Theoretical Yield D.Percent Error

?2 $500 This 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 chemistry B. Conservation chemistry C. Stoichiometry D. Quantum mechanics

This is the volume of O 2 formed when 3.50 moles of water decomposes to hydrogen and oxygen in the following equation: 2 H 2 O → 2 H 2 + O 2 A L O2 B L H2 C L O2 D L H2 ?3 $100

A3 $100

This is the number of grams of potassium chloride formed when 27.8 g of potassium chlorate reacts in the following equation: 2 KClO 3 → 2 KCl + 3 O 2 A g KCl B g KCl C g KCl D g KCl ?3 $200

A3 $200

?3 $300 This is the number of moles of H 2 O formed when 33.0 grams of H 2 SO 4 reacts with KOH. H 2 SO KOH → K 2 SO H 2 O A moles H2O B moles H2O C moles H2O D moles H2O

A3 $300

This is the theoretical yield of sodium chloride when 19.4 g of sodium react with chlorine in the following equation: 2 Na + Cl 2 → 2 NaCl A g B g C g D g ?3 $400

A3 $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 O 2 → 2 Fe 2 O 3 Lab Data Mass of iron reacted: 67.4 g Mass of iron (III) oxide formed: g A. Law of Conservation of Mass B. Percent yield is 100% C. Percent error is 0% D. B & C ?3 $500

A3 $500 What is 46.5 g of O 2 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 number of phosphorous atoms in Mg 3 (PO 4 ) 2 A. 1 B. 2 C. 6 D. 8 ?4 $100

This is the formula in the list below which is not an empirical formula A. K 2 S B. C 4 H 10 C. Li 2 SO 4 D. H 3 PO 4 ?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 % B % C % D. 125 % ?4 $300

This is the percent composition of sulfur in H 2 SO 4 A % B % C % D % ?4 $400

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 - C 4 H 6 O? 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 + O 2 → 2MgO Data Table A g B g ?5 $100 Constant mass of evaporating dish27.88 g Mass of evaporating dish and Mg30.14 g Mass of evaporating dish and MgO (first heating) g Mass of evaporating dish and MgO (second heating) g Constant mass of evaporating dish and MgO g

This is the experimental yield of MgO from the following reaction: 2Mg + O 2 → 2MgO Data Table A g B g ?5 $200 Constant mass of evaporating dish27.88 g Mass of evaporating dish and Mg30.14 g Mass of evaporating dish and MgO (first heating) g Mass of evaporating dish and MgO (second heating) g Constant mass of evaporating dish and MgO g

This is a reason why the theoretical ratio of a hydrate is often lower than the experimental ratio. A. Humidity in the environment B. Insufficient heating C. Calculator error D. A & B ?5 $300

These are reasons why experimental yield is often less than theoretical yield. A. The reaction does not go to completion B. Product may be lost due to poor technique (spilling, transfer, etc.) C. Reactants may not be pure D. Side reactions may occur E. All of the above ?5 $400

Calculate the percent yield of the equation below using the information in the data table. 2Mg + O 2 → 2MgO A % B % ?5 $500 Constant mass of evaporating dish31.95 g Mass of evaporating dish and Mg35.12 g Mass of evaporating dish and MgO (first heating) g Mass of evaporating dish and MgO (second heating) g Constant mass of evaporating dish and MgO36.74 g

Mass of reactant = – = 3.17 g Mg Theoretical Yield = 5.26 g MgO Actual Yield = – = 4.79 g MgO Percent Yield = 4.79/5.26 * 100 = 91.1% Constant mass of evaporating dish31.95 g Mass of evaporating dish and Mg35.12 g Mass of evaporating dish and MgO (first heating) g Mass of evaporating dish and MgO (second heating) g Constant mass of evaporating dish and MgO36.74 g