CHEM Quiz Answer Key

Quiz 1 Chemistry 123 Harwood/Fall 05 KEY Quiz 1A 1. If you burn 1.959 g of Fex(CO)y and find 0.799 g of Fe2O3 and 2.200 g of CO2, what is the empirical formula of Fex(CO) y? (3 pts) (a) Fe(CO)4 (b) Fe2(CO)9 ! 1 mol Fe2O 3 $ ! 2 mol Fe $ ! 55.8477 g $ ! 1 mol Fe $ mol Fe = ( 0.799 g Fe2O 3 ) # = 0.010008 mol Fe ' 1 # & " 159.6922 g &% #" 1 mol Fe2O 3 &% " 1 mol Fe % #" 55.847 g &% (c) Fe(CO)5 ! 1 mol CO2 $ ! 1 mol C $ ! 12.011 g $ ! 1 mol C $ mol C = ( 2.200 g CO2 ) # = 0.04999 mol C ' 5 # & " 44.0098 g &% #" 1 mol CO2 &% " 1 mol C % #" 12.011 g &% (d) Fe(CO)6 ! 1 mol O $ mol O = [1.959 g - (0.5589 g + 0.6004 g)] # = 0.04998 mol O ' 5 " 15.9994 g &% 2. In an experiment, 1.056 g of a metal carbonate containing an unknown metal M was heated to give the metal oxide and 0.376 g CO2. What is the identity of the metal M? (2 pts) (a) M = Ni (b) M = Cu (c) M = Zn (d) M = Ba ! 1 mol CO2 $ mol CO2 = ( 0.376 g CO2 ) # = 0.008544 mol CO2 " 44.0098 g &% ! 1 mol MCO 3 $ mol MCO 3 = ( 0.008544 g CO2 ) # = 0.008544 mol MCO 3 " 1 mol CO2 &% 1.056 g = 123.602 g/mol 0.008544 mol atomic mass of M = 123.602 g/mol - [12.011 + (3 x 15.994)] = 63.59 g/mol molar mass MCO 3 = – – – 3. A new element, lubine (Lb), has been discovered. It forms four oxoanions: LbO2 , LbO3 , LbO4 , – LbO5 . Name the following compounds: (2 pts) (a) NaLbO3 sodium lubite (b) NaLbO5 sodium perlubate 4. Identify the type of chemical reaction represented by the following equations. (3 pts) (a) 3 H2 (g) + N2 (g) → 2 NH3 (g) combination (b) AgNO3 (aq) + NaCl (aq) → NaNO3 (aq) + AgCl (s) exchange (c) CaCO3 (s) → CaO (s) + CO2 (g) decomposition Quiz 1B 1. Same as 1A 2. Same as 1A – – – 3. A new element, lubine (Lb), has been discovered. It forms four oxoanions: LbO2 , LbO3 , LbO4 , – LbO5 . Name the following compounds: (2 pts) (a) NaLbO2 sodium hypolubite (b) NaLbO4 sodium lubate 4. Identify the type of chemical reaction represented by the following equations. (a) P4O10 (s) + 6 H 2O (l) → 4 H3PO4 (l) combination (b) 2 Al (s) + 3 H2SO4 (aq) → Al2(SO4)3 (s) + 3 H2 (g) displacement (c) 2 Na (s) + 2 H2O (l)→ 2 NaOH (aq) + H2 (g) displacement (3 pts) Quiz 1A Chemistry 123 Harwood/Fall 05 Quiz 1C 1. The complete reaction of 3.84 g of vanadium with 8.16 g of chlorine (Cl2) produces a compound with the formula VxCly. What is the empirical formula of the compound? (3 pts) (a) VCl (b) VCl2 (c) VCl3 ! 1 mol V $ mol V = ( 3.84 g V ) # = 0.07538 mol V ' 1 " 50.9415 g &% ! 1 mol Cl2 $ ! 2 mol Cl $ mol Cl = ( 8.16 g Cl2 ) # = 0.2302 mol Cl ' 3 " 70.906 g &% #" 1 mol Cl2 &% (d) VCl4 (e) VCl5 2. In an experiment, 1.056 g of a metal carbonate containing an unknown metal M was heated to give the metal oxide and 0.391 g CO2. What is the identity of the metal M? (2 pts) (a) M = Ni (b) M = Cu (c) M = Zn (d) M = Ba ! 1 mol CO2 $ mol CO2 = ( 0.391 g CO2 ) # = 0.008884 mol CO2 " 44.0098 g &% ! 1 mol MCO 3 $ mol MCO 3 = ( 0.008884 g CO2 ) # = 0.008884 mol MCO 3 " 1 mol CO2 &% 1.056 g = 118.860 g/mol 0.008884 mol atomic mass of M = 118.860 g/mol - [12.011 + (3 x 15.9994)] = 58.85 g/mol molar mass MCO 3 = – – – 3. A new element, lubine (Lb), has been discovered. It forms four oxoanions: LbO2 , LbO3 , LbO4 , – LbO5 . Name the following compounds: (2 pts) (a) Ca(LbO2)2 calcium hypolubite (b) Ca(LbO4)2 calcium lubate 4. Identify the type of chemical reaction represented by the following equations. (3 pts) (a) 3 Fe (s) + 4 H2O (g)  Fe3O4 (s) + 4 H2 (g) displacement (b) 2 AgNO3 (aq) + BaCl2 (aq) → Ba(NO3)2 (aq) + 2 AgCl (s) exchange (c) 2 KClO3 (s)  2 KCl (s) + 3 O2 (g) decomposition Quiz 1D 1. Same as 1C 2. Same as 1C – – – 3. A new element, lubine (Lb), has been discovered. It forms four oxoanions: LbO2 , LbO3 , LbO4 , – LbO5 . Name the following compounds: (2 pts) (a) Ca(LbO2)2 calcium lubite (b) Ca(LbO5)2 calcium perlubate 4. Identify the type of chemical reaction represented by the following equations. (a) 2 Al (s) + 3 CuSO4 (aq) → Al2(SO4)3 (aq) + 3 Cu (s) displacement (b) MgSO3 (s) → MgO (s) + SO2 (g) decomposition (c) 2 H2O (l) → 2 H2 (g) + O2 (g) decomposition (3 pts)