03/22/16
CHEM 4
ANSWER KEY
Name _________________________________
Quiz 6 (20 points)
1. (5 points) Fill the blanks in the following table and answer the additional question below the
table.
Balanced Chemical
Equation
Moles before the reaction
N2(g)
+
3 H2(g)
→
2 NH3(g)
3.0 mol
6.0 mol
0 mol
Moles consumed (–) or
moles produced (+)
− 0.5 mol
− 1.5 mol
+ 1.0 mol
Moles after the reaction
2.5 mol
4.5 mol
1.0 mol
What is the percent yield for the reaction from the above table? Show work.
H2 is the limiting reactant.
% yield = (1.5 mol H2) / (6.0 mol H2) × 100% = 25%
2. (5points) Use the molecular diagram given below to answer the questions that follow.
Write the balanced chemical equation for the reaction schematically depicted in the diagram.
2 NH3 + 3 N2O → 4 N2 + 3 H2O
Which substance is the limiting reactant? What is the percent yield of the reaction? Briefly
explain.
N2O is completely used in the reaction. Therefore, N2O is the limiting reactant
and the percent yield is 100%.
PLEASE TURN OVER!!!
3. (10 points) Write the full-formula, complete ionic, and net ionic equation (FFE, CIE, and
NIE) for each of the two reactions described below. In each equation, indicate physical state
of each reactant and product: (s), (l), (g), or (aq).
(a) Aqueous solutions of barium nitrate and phosphoric acid are mixed.
FFE: 3 Ba(NO3)2(aq) + 2 H3PO4(aq) → Ba3(PO4)2(s) + 6 HNO3(aq)
CIE: 3 Ba2+(aq) + 6 NO3−(aq) + 2 H3PO4(aq) → Ba3(PO4)2(s) + 6 H+(aq) + 6 NO3−(aq)
NIE: 3 Ba2+(aq) + 2 H3PO4(aq) → Ba3(PO4)2(s) + 6 H+(aq)
(b) Aluminum metal is dissolved in sulfuric acid.
FFE: 2 Al(s) + 3 H2SO4(aq) → Al2(SO4)3(aq) + 3 H2(g)
CIE: 2 Al(s) + 6 H+(aq) + 3 SO42−(aq) → 2 Al3+(aq) + 3 SO42−(aq) + 3 H2(g)
NIE: 2 Al(s) + 6 H+(aq) → 2 Al3+(aq) + 3 H2(g)