S2i
Chemistry 212
Sunday, September 30 th
Katie Seehusen
Kseehusen09@winona.edu
Agenda:
1.
Limiting Reactants
True or False? If false, correct the sentence.
T / F The limiting reactant is the reactant that produces the most product.
______________________________________________the least________
T / F To compare two compounds to each other using the coefficients of the equation, I need to put them into moles.
____________________________________________________________
3.76
Which reagent is the limiting reactant when 0.450 mol Al(OH)
3
H
2
SO
4
are allowed to react?
and 0.450 mol
.45 mol Al(OH)
3
1 mol Al
2
(SO
4
)
3
2 mol Al(OH)
3
= 0.225 mol Al
2
(SO
4
)
3
= 0.15 mol Al
2
(SO
4
)
3
***limiting reactant .45 mol H
2
SO
4
1 mol Al
2
(SO
4
)
3
3 mol H
2
SO
4

How many moles of Al
2
(SO
4
)
3
will be produced?
0.15 mol (amount produced using 0.45 mol H
2
SO
4
How many moles of the excess reactant remain?
Amount of non-limiting reactant to start - amount of non-limiting reactant used
0.45 mol H
2
SO
4
2 mol Al(OH)
3
= 0.30 mol Al(OH)
3
3 mol H
2
SO
4
0.45 mol Al(OH)
3
– 0.3 mol Al(OH)
3
= 0.15 mol Al(OH)
3
left over
3.82
When ethane (C
2
H
6
) reacts with chlorine (Cl
2
), the main product is C
2
H
5
Cl; but other products also are produced.
Calculate the theoretical yield of C
2
H
5
Cl when 124 g of C
2
H
6
reacts with 232 g of Cl
2 assuming the only products formed are C
2
H
5
Cl and HCl.
125 g C
2
H
6
1 mol C
2
H
30.068 g C
6
2
H
6
1 mol C
1 mol C
2
2
H
H
5
6
Cl 64.51 g C
1 mol C
2
2
H
H
5
5
Cl
Cl
= 268.2 g
C
2
H
5
Cl
232 g Cl
2
1 mol Cl
70.9 g Cl
2
2
1 mol C
2
H
1 mol Cl
2
5
Cl 64.51 g
C
2
H
5
Cl
1 mol C
2
H
5
Cl
= 211.1 g C yield
2
H
5
Cl
**limiting reactant produces theoretical

3.83
If you started with 45.0 grams of H
2
S and 50.0 grams of O
2
, how many grams of S would be produced, assuming 95% yield?
8
45 g H
2
S 1 mol H
2
S 1 mol S
8
256.56 g S
8
= 42.3 g S
8
**limiting
34.086 g H
2
S 8 mol H
2
S 1 mol S
8
50 g O
2
1 mol O
2
32 g O
2
1 mol S
4 mol O
8
2
256.56 g S
1 mol S
8
8
= 100.2 g S
8
Theoretical Yield: 42.3 g (0.95)= 40.18 g S
8
3.84
When hydrogen sulfide gas is bubbled into a solution of sodium hydroxide, the reaction forms sodium sulfide and water. How many grams of sodium sulfide are formed if 1.60 g of hydrogen sulfide is bubbled into a solution containing 2.13 g of sodium hydroxide, assuming that the sodium sulfide is made in 94.0% yield?
H
2
S + 2 NaOH è Na
2
S + 2 H
2
O
1.6 g H
2
S 1 mol H
2
S
34.086 g H
2
1 mol Na
2
S 1 mol H
2
S
S
78.05 g Na
2
S
1 mol Na
2
S
= 3.66 g Na
2
S
2.13 g NaOH 1 mol NaOH 1 mol Na
2
S 78.05 g Na
2
S
40 g NaOH 2 mol NaOH 1 mol Na
2
S
= 2.08 g Na
**limiting
2
S
2.08g Na
2
S (0.94)= 1.95 g Na
2
S theoretical yield