1) In the combustion analysis of 0.1127 g of glucose (C6H12O6), what mass, in grams, of CO2 would be produced?
A. 0.0451 g
B. 0.0825 g
C. 0.1652 g The solution is C because you have to write the combustion out as an equation then once the equation is balanced then convert the mass of the glucose and convert it to the CO2 mass in grams.
D. 0.4132 g
E. 1.466 g
2) Sulfur dioxide reacts with chlorine to produce thionyl chloride (used as a drying agent for inorganic halides) and dichlorine oxide (used as a bleach for wood, pulp and textiles).

SO2(g) + 2Cl2(g) → SOCl2(g) + Cl2O(g)

If 0.400 mol of Cl2 reacts with excess SO2, how many moles of Cl2O are formed?
A. 0.800 mol
B. 0.400 mol Take the moles value that was given and then convert it into the Cl2O by dividing .400 by the mole value 2 to cancel out the mole value infront of the Cl2 in the original equation.
C. 0.200 mol
D. 0.100 mol
E. 0.0500 mol

3) Aluminum oxide (used as an adsorbent or a catalyst for organic reactions) forms when aluminum reacts with oxygen.

4Al(s) + 3O2(g) → 2Al2O3(s)

A mixture of 82.49 g of aluminum ( = 26.98 g/mol) and 117.65 g of oxygen ( = 32.00 g/mol) is allowed to react. What mass of aluminum oxide ( = 101.96 g/mol) can be formed?
A. 155.8 g
B. 200.2 g
C. 249.9 g Convert the masses of aluminum to Aluminum oxide and do the same for the Oxygen then once that was completed then find the mass value of Aluminum oxide that contains the lowest value (limiting reagent).
D. 311.7 g

E. 374.9 g
4) Magnesium (used in the manufacture of light alloys) reacts with iron(III) chloride to form magnesium chloride and iron.

3Mg(s) + 2FeCl3(s) → 3MgCl2(s) + 2Fe(s)

A mixture of 41.0 g of magnesium ( = 24.31 g/mol) and 175 g of iron(III) chloride ( = 162.2 g/mol) is allowed to react. Identify the limiting reactant and determine the mass of the excess reactant present in the vessel when the reaction is complete.
A. Limiting reactant is Mg; 67 g of FeCl3 remain.
B. Limiting reactant is Mg; 134 g of FeCl3 remain.
C. Limiting reactant is Mg; 104 g of FeCl3 remain. Convert the masses of the Magnesium and iron(III) chloride into the mass of one of the products that were produce and then the one that produces that least amount would be the (limiting reagent) then by subtracting that from the higher value would result in the excess value.
D. Limiting reactant is FeCl3; 3 g of Mg remain.
E. Limiting reactant is FeCl3; 87 g of Mg remain.
5) Tetraphosphorus hexaoxide ( = 219.9 g/mol) is formed by the reaction of phosphorus with oxygen gas.

P4(s) + 3O2(g) → P4O6(s)

If a mixture of 75.3 g of phosphorus and 38.7 g of oxygen produce 43.3 g of P4O6, what is the percent yield for the reaction?
A. 57.5%
B. 48.8%
C. 38.0% Find the mass grams of the final product and then using the different masses given from the different reactants then the one that would be the limitiing reactant and then divide the mass of the acutal which was 43.3g and then divide by the limiting.
D. 32.4%
E. 16.3%
6) Methanol (CH4O) is converted to bromomethane (CH3Br) as follows:

CH4O + HBr → CH3Br + H2O

If 12.23 g of bromomethane are produced when 5.00 g of methanol is reacted with excess HBr, what is the percentage yield?
A. 40.9%
B. 82.6% This would be the same as the problem above find the limiting reactance and then divide by the actual from the mass of the limiting reactant.
C. 100.%
D. 121%
E. 245%
7) A 0.150 M sodium chloride solution is referred to as a physiological saline solution because it has the same concentration of salts as normal human blood. Calculate the mass of solute needed to prepare 275.0 mL of a physiological saline solution.
A. 41.3 g
B. 31.9 g
C. 16.1 g find the mass of the solute by multiplying .150 by 1L because M=(mol/L) then multiply by 275 mL and divide by 1000mL and then take that mol value and convert into the grams value.
D. 8.77 g
E. 2.41 g
8) Copper(II) sulfide, CuS, is used in the development of aniline black dye in textile printing. What is the maximum mass of CuS which can be formed when 38.0 mL of 0.500 M CuCl2 are mixed with 42.0 mL of 0.600 M (NH4)2S? Aqueous ammonium chloride is the other product.
A. 2.41 g
B. 1.82 g
C. 1.21 g
D. 0.909 g
E. 0.044 g
9) Aluminum will react with bromine to form aluminum bromide (used as an acid catalyst in organic synthesis).

Al(s) + Br2(l) → Al2Br6(s) [unbalanced]

How many moles of Al are needed to form 2.43 mol of Al2Br6?
A. 7.29 mol
B. 4.86 mol
C. 2.43 mol
D. 1.62 mol
E. 1.22 mol
10) How many grams of sodium fluoride (used in water fluoridation and manufacture of insecticides) are needed to form 485 g of sulfur tetrafluoride?

3SCl2(l) + 4NaF(s) → SF4(g) + S2Cl2(l) + 4NaCl(s)
A. 1940 g
B. 1510 g
C. 754 g
D. 205 g
E. 51.3 g
11) Magnesium reacts with iron(III) chloride to form magnesium chloride (which can be used in fireproofing wood and in disinfectants) and iron.

3Mg(s) + 2FeCl3(s) → 3MgCl2(s) + 2Fe(s)

A mixture of 41.0 g of magnesium ( = 24.31 g/mol) and 175 g of iron(III) chloride ( = 162.2 g/mol) is allowed to react. What mass of magnesium chloride = 95.21 g/mol) is formed?
A. 68.5 g MgCl2
B. 77.0 g MgCl2
C. 71.4 g MgCl2
D. 107 g MgCl2
E. 154 g MgCl2
12) Potassium chloride is used as a substitute for sodium chloride for individuals with high blood pressure. Identify the limiting reactant and determine the mass of the excess reactant remaining when 7.00 g of chlorine gas reacts with 5.00 g of potassium to form potassium chloride.
A. Potassium is the limiting reactant; 2.47 g of chlorine remain.
B. Potassium is the limiting reactant; 7.23 g of chlorine remain.
C. Chlorine is the limiting reactant; 4.64 g of potassium remain.

D. Chlorine is the limiting reactant; 2.70 g of potassium remain.
E. No limiting reagent: the reactants are present in the correct stoichiometric ratio.
13) Hydrochloric acid is widely used as a laboratory reagent, in refining ore for the production of tin and tantalum, and as a catalyst in organic reactions. Calculate the number of moles of HCl in 62.85 mL of 0.453 M hydrochloric acid.
A. 28.5 mol
B. 1.04 mol Notes:
This unit has g:mol:g ratio with a reaction between Products and the Reactants. This also has a balanced equations to the the relationship and compounds. Mole calculations has finding a mol mass using the original mass of the reactants. Molarity is also involved which includes the M(mol/L) inorder to find it requires converting mL and grams into the Molarity using factor labeling. Also, there was a limiting reagent which is acutally doing the mol:g:mol ratio of a balanced equation and the one which gives a product with the smallest mass is the limiting and the other the excess. When subtracting the limiting rectanct from the excess results in the in the remaining mass of a substance. There is also a percent yield which is calculated by the actual value (starting value) - the theoretical value that you find by using stoichemostry of the reaction. That is mainly what is stoic is.
M(mol/L), g:mol:g
C. 0.139 mol
D. 0.0285 mol
E. 0.00721 mol