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Key for test 2 guideline sample questions:
- answer: #g Na 2
CO
needed=
.250L(.14molNa/L)(1molNa 2
CO
/2molNa)(106g/mol)=1.86g
- a) sodium sulfate & barium chloride. React? yes (yes/no) product(s): BaSO4 ppt
& NaCl(aq)
BaCl 2 (aq)+ Na 2
SO
4 (aq) โโ> BaSO 4 (s) + 2 NaCl ; net: Ba
2+
2 โ โโ> BaSO 4 (s)
b) silver chloride & potassium nitrate. React? no (yes/no) If so, product(s):
None since AgCl is insoluble and will not be able to go into solution.
c) Lead acetate & ammonium iodide. React? yes (yes/no) If so, product(s): PbI 2
Pb(CH 3
CO
2
2 (aq) + 2NH 4 I (aq) ๏ PbI 2 (s) + 2 NH 4
CH
3
CO
2 (aq).
Net ionic: Pb
2+
โ โโ>PbI 2
d) sulfuric acid + sodium hydrogen carbonate? yes (yes/no) If so, product(s):
Na 2
SO
4
, CO
2 (g), H 2 O(l).
H
2
SO
4 (aq) + 2 NaHCO 3 (aq) โโ> Na 2
SO
4 (aq) + 2 CO 2 (g) + 2 H 2 O (l)
3 ) a) H 3
PO
โโ> Na 3
PO
+ 3H
CO
b) (NH 4
SO
โโ> BaSO 4
+ 2NH
NO
MnO 4
โ
โ โโ> Mn
2+
O
OK, now although the Oโs are balanced, a new problem arises, we have added 8 Hโs
on the right side. To balance those, we add 8 H
= โs on the left side:
8H
โ
โ โโ> Mn
2+
- 4H 2 O (finally this half reaction is balanced!)
finally, check to see if the charges are balanced. Yes they are! Both sides have a net
+2 charge.
Now, we have to add the 2 half reactions. But not before making sure that the
electrons will cancel out. There must absolutely be no net eโs in the final balanced
equation. By inspection we note that this will only happen if we first multiply the
first (oxidation) equation by 5 and the second (reduction) equation by 2, prior to
adding both up.
Final balanced REDOX equation:
5C
2
O
4
2 โ
โ โโ> 10CO 2
2+
- 8H 2 O (Balanced by charge
and by mass! Count the charges and the atoms)
4 ) a) It is a good habit to start by inspecting the reaction by writing the balanced
equation : 2 NaOH + H 2
SO
โโ> 2H
O + Na 2
SO
This is a titration problem. If the coefficients were both 1, it would be a simple
matter of using M 1
V
1
=M
2
V
2
. But it is not so simple, so we can just treat it as a
standard stoichiometry problem. At the equivalence point, we have:
[NaOH]=moles of NaOH/L NaOH = moles H SO (2 mol NaOH/mol H SO ) / L NaOH
but moles H 2
SO
4
= [H
2
SO
4 ] x (L H 2
SO
4 ) = (0.25mol/L)(0.0150L). so, we can
write the full solution below:
.015(.25) mol H 2
SO
(2mol NaOH /mol H 2
SO
)/0.025L = 0.300 M NaOH.
(sometimes people get confused about what this means. It means that initially,
before we even started any titration, the concentration of the NaOH was 0.300M.
Titrations are of course done to determine the otherwise unknown concentration of
a base or acid solution).
Here is an alternative approach: (this how we reason it out in Chem 201)
At equivalence: # equiv H 2
SO
4 = # equiv NaOH
2 x #moles H 2
SO
4 = #moles NaOH
But moles = MV, so, we can write:
2 x M H2SO
V
H2SO
= M
NaOH
V
NaOH
=> 2M
1
V
1
= M
2
V
e where V e = vol
at equiv
M
2
=2M
1
V
1
/V
e =2(15mL)(.250M)/(25 mL)= 0.300M (note that
the volume is kept as โmLsโ because the units cancel out in the end anyway.)
b) In this question, we have NOT YET REACHED EQUIVALENCE. So, we first
calculate how many NaOHโs have reacted based on the H 2
SO
4 as the limiting
reagent. Then, we calculate how much NaOH has remained unreacted by just
subtracting it off from the original NaOH. For this problem, it is safest to think in
terms of MOLES. Note that this problem is asking what the concentration of
NaOH is during the titration. Thus, we need to take into account the actual total
volume of the solution after adding H 2
SO
4 to it.
Moles NaOH reacted = moles H 2
SO
4 added x (2 mol NaOH/1mol H 2
SO
4 added)
= (.250 mol H 2
SO
4
/L)(.0100L)(2)
the abbreviated solution looks like this:
[NaOH]=mol NaOH left over
/L total soln = mol NaOH i
โmol NaOH reacted
)/(vol
NaOH i
+vol H 2
SO
4 added
) = (M
V
โ2M
V
)/(V
+V
2(.25)(10))/(25+10)=.0714 M NaOH
5 ) (a) Note that the number of moles of H 2
SO
4 is determined by using conversion
factors. That is then divided by the total final volume (in liters).
#mol N 2
needed = 12.0 mol H 2
x (1 mol N 2
/3mol H 2
)= 4.0 mol N 2
. Since that is less
than the 6.0 mol available, N 2
must be in excess by 2.0 mols. H 2
is limiting
reactant. #g N 2
excess = 2.0 mol(28.0g/mol)=56.0 g N 2
b) Theoretical yield of NH 3
= (12.0 mol H 2
)(2 mol NH 3
/3 mol H 2
) =8.0 mol NH 3
(17.0g/mol)=136 g NH 3
c) Recall that % yield = (actual yield/theoretical yield)x100%
=> actual yield = (%yield/100%)(theoretical yield)
So, # mol NH 3
= (0.80)( 8.0 mol)=6.4 mol.
d) #g NH 3
=6.4 mol.x(17.0g/mol)=110 g NH 3
e) # molecules NH 3 = 6.4 mol NH 3 x (6.02x
23 molec/mol)= 3.85x
24 molecules
f)V NH = 110g(1L/0.76g)= 140 L
7 ) Write balanced equation for this reaction:
Cu(s) + 4HNO 3
โโ> Cu(NO 3
+ 2NO
(g)+ 2H 2
O
This is a conversion factor problem:
#L NO
= (0.15cm
3 Cu)(8.95g Cu /cm
3 Cu)(1mol Cu /63.5g
Cu)(2molNO 2
/molCu) (1L NO 2
/2.05g NO 2
)(0.70) = 0.014L
8 ) recall MV=moles; so MV of H 2
C
O
gives moles H 2
C
O
. From there convert to
moles KOH then concentration. But you need balanced eqn: [KOH]=?
H
C
O
+ 2KOH โโ> K
C
O
+ 2H
O:
[H
C
O
]=.0200L KOH (.400molKOH/L KOH)(1mol H 2
C
O
/mol KOH)/.020L
H
C
O
= 0.25 M H
C
O
alternatively, M 1
V
1
= 2 M
2
V
2 (which is โ1โ? โ1โ is KOH in this equation. If you
canโt clearly determine which is which, donโt use this alternative approach. You
will have only a 50% chance of succeeding).
M
2
=M
1
V
1
/(2V
2 ) = (.400M)(20.0mL)/(2(16.0mL) = 0.25 M H C O
