1) For the following synthesis/combustion/Redox reaction:
2 Sr (s) + O2 (g) -----> 2 SrO (s)
How many grams of metal oxide are formed when 0.494 mol of strontium react with excess oxygen?
0.494 mol Sr x 2 mol SrO x 103.62 g SrO
2 mol Sr
1 mol SrO
= 51.1¦8828 g SrO ------> 51.2 g SrO
2) For the following precipitation/double displacement reaction:
3 Zn(ClO3)2 (aq) + 2 K3PO4 (aq) ------> Zn3(PO4)2 (s) + 6 KClO3 (aq)
How many grams of potassium phosphate are required to form 12.7 g of precipitate?
12.7 g Zn3(PO4)2 x 1 mol Zn3(PO4)2
386.1 g Zn3(PO4)2
x
2 mol K3PO4
1 mol Zn3(PO4)2
x 212.266 g K3PO4 = 13.9¦641 g
1 mol K3PO4
----> 14.0 g K3PO4
3) For the following neutralization reaction:
H3O+ (aq) + Cl– (aq) + Na+ (aq) + OH– (aq) -------> 2 H2O (l) + Na+ (aq) + Cl– (aq)
How many grams of water are formed when 0.998 g of hydronium ions react with excess base?
H3O+ molar mass:
H2O molar mass:
0.998 g H3O+ x
(3 x 1.00794) + (1 x 15.9994) = 3.02382 + 15.9994 = 19.0232¦2 ------> 19.0232 g/mol
(2 x 1.00794) + (1 x 15.9994) = 2.01588 + 15.9994 = 18.0152¦8 ------> 18.0153 g/mol
1 mol H3O+ x 2 mol H2O x 18.0153 g H2O = 1.89¦0246 ----> 1.89 g H2O
19.0232 g H3O+
1 mol H3O+
1 mol H2O
4) How many mg of precipitate are formed when reacting 12.05 g of barium chloride in the following:
2 AgNO3 (aq) + BaCl2 (aq) ------> 2 AgCl (s) + Ba(NO3)2 (aq)
12.05 g BaCl2 x 1 mol BaCl2 x 2 mol AgCl x 143.3209 g AgCl x 1000 mg AgCl = 16,58¦6.79 mg AgCl
208.24 g BaCl2 1 mol BaCl2
1 mol AgCl
1g AgCl -----> 1.659 x 104 mg AgCl
5) How many mg of Cr(NO3)3 (aq) are required to completely react with 0.433 g of (NH4)2S (aq)?
2 Cr(NO3)3 (aq) + 3 (NH4)2S (aq) -------> 6 NH4NO3 (aq) + Cr2S3 (s)
0.433 g of (NH4)2S x
1 mol (NH4)2S
x
68.143 g (NH4)2S
2 mol Cr(NO3)3 x 238.011 g Cr(NO3)3 x 1000 mg
3 mol (NH4)2S
1 mol Cr(NO3)3
1g
= 100¦8.2597 mg -------> 1.01x103 mg Cr(NO3)3
6)
Laughing gas, N2O, is made by the careful thermal decomposition of ammonium nitrate.
NH4NO3(s) Æ N2O(g) +
a.
2H2O(l)
If you begin with 1.00x103 g of ammonium nitrate, how many grams of laughing gas can you
obtain?
(1.00x103 g NH4NO3) * (1 mol NH4NO3) * (1 mol N2O) * (44.02¦ g N2O) = …
(80.05¦2 g NH4NO3) (1 mol NH4NO3) (1 mol N2O)
= 549.¦892 g N2O = 5.50x102 g N2O
b.
If you begin with 1.00x103 g of ammonium nitrate, how many grams of water can you obtain?
(1.00x103 g NH4NO3) * (1 mol NH4NO3) * (2 mol H2O) * (18.01¦6 g H2O) = …
(80.05¦2 g NH4NO3) (1 mol NH4NO3) (1 mol H2O)
= 450.¦107 g H2O = 4.50x102 g H2O
7)
Li2CO3 (aq) +
Ca(C2H3O2)2 (aq) ------> 2 LiC2H3O2 (aq) + CaCO3 (s)
(a) If you started with 2.99 g of Li2CO3 and an unlimited supply of Ca(C2H3O2)2, how many
grams of CaCO3 could be produced?:
2.99 g Li2CO3 ×
1 mol Li2CO3 1 mol CaCO3 100.087 g CaCO3
×
×
= 4.05 | 01 g CaCO3
73.89 g Li2CO3 1 mol Li2CO3
1 mol CaCO3
(b) If you started with 4.077 g of Ca(C2H3O2)2 and an unlimited supply of Li2CO3, how many
grams of CaCO3 could be produced?
4.077 g Ca(C2 H 3O 2 ) 2 ×
1 mol Ca(C2 H 3O 2 ) 2
1 mol CaCO3
100.087 g CaCO3
×
×
= 2.579 | 90 g CaCO3
158.167 g Ca(C2 H 3O 2 ) 2 1 mol Ca(C2 H 3O 2 ) 2
1 mol CaCO3
(c) Now assume you added the 2.99 g of Li2CO3 to the 4.077 g of Ca(C2H3O2)2, what mass of
CaCO3 could be produced? You do not need to do any calculations for this question. Look at your
solutions to parts a) and b) above and determine your answer.
Now calculate how many grams of ppt are formed:
In the calculations above, we calculated the amount of CaCO3 that could be produced if Li2CO3 was
the limiting reactant and also the amount of CaCO3 that could be produced if Ca(C2H3O2)2 was the
limiting reactant. If all of the Ca(C2H3O2)2 reacts, it would produce produces less CaCO3 than if all
of the Li2CO3 reacted. Therefore the Ca(C2H3O2)2 is the limiting reactant and 2.580 g CaCO3 will be
produced.
8) Calculate how many tons of CO2 (g) you release in one year driving. Assume the density of gasoline (use
octane) to be 0.814 g/ml:
2 C8H18 (g) + 25 O2 (g) -----> 16 CO2 (g) + 18 H2O (g) combustion of gasoline (as octane)
(18,000 miles)*( 1 gal) * (3.7854 L)*(1000 mL) * (0.814 g C8H16) = 2521076 g C8H18
(year)
(22 miles) (1 gallon) (1L)
(mL)
= 2.5 x 106 g C8H18
For mass of CO2 formed:
C8H18 molar mass: (8 x 12.011) + (18 x 1.00794) = 96.088 + 18.1429¦2 = 114.230¦92 -----> 114.231 g/mol
We will use 25|21076 g C8H18 in the next calculation, but remember that it will limit us to two significant figures…
(25|21076 g C8H18)*(1mol C8H18) (16 mol CO2) (44.010 g CO2)(1 lb ) (1 ton) = 8.5¦65 tons CO2
(114.231 g C8H18)(2 mol C8H18) (1 mol CO2) (453.59g)(2000 lb)
= 8.6 tons CO2
9)
Methyl alcohol (wood alcohol), CH3OH, is produced via the reaction
CO(g) + 2 H2(g) → CH3OH(l)
A mixture of 15.21 g H2(g) and 54.52 g CO(g) are allowed to react.
(a) Which reagent is the limiting reagent?
(54.52 g CO) * (1 mol CO) * (1 mol CH3OH) * (32.04¦2 g CH3OH) =
(28.01¦ g CO) (1 mol CO)
(1 mol CH3OH)
= 62.36¦80 g CH3OH = 62.37 g CH3OH
(15.21 g H2) * (1 mol H2) (1 mol CH3OH) (32.04¦2 g CH3OH) =
(2.016 g H2) (2 mol H2)
(1 mol CH3OH)
= 120.8¦727 g CH3OH = 120.9 g CH3OH
Therefore, CO is the limiting reactant. (and H2 is in excess)
(b) What is the theoretical yield of CH3OH in grams?
62.37 g CH3OH
(c) How many grams of the reagent present in excess is left over?
(62.36¦80 g CH3OH) * (1 mol CH3OH) * (2 mol H2) * (2.016 g H2) = 7.847¦06 g H2 required to react
(32.04¦2 g CH3OH) (1 mol CH3OH) (1 mol H2)
EXCESS: 15.21 g H2 – 7.847¦06 g H2 = 7.36¦29 g H2 = 7.36 g H2
(d) Suppose the actual yield is 22.52 g of CH3OH. What is the % yield?
(22.52 g /62.36¦80 g)*100% = 36.10¦826 % yield = 36.11% yield
10)
Given the following equation:
Al2(SO3)3 + 6 NaOH ------> 3 Na2SO3 + 2 Al(OH)3
a)
If 10.00 g of Al2(SO3)3 is reacted with 10.00 g of NaOH, determine the limiting reagent
(10.00 g Al2(SO3)3) * (1 mol Al2(SO3)3) *(3 mol Na2SO3) *(126.044¦ g Na2SO3) = 12.85¦42 g Na2SO3
(294.17¦ g Al2(SO3)3) (1 mol Al2(SO3)3) (1 mole Na2SO3)
(10.00 g NaOH) * (1 mol NaOH) *(3 mol Na2SO3) *(126.044¦ g Na2SO3) = 15.75¦63 g Na2SO3
(39.99¦8 g NaOH) (6 mol NaOH)
(1 mole Na2SO3)
Hence, Al2(SO3)3 is the limiting reactant. (NaOH is in excess)
b) Determine the number of moles of Al(OH)3 produced
(10.00 g Al2(SO3)3) (1 mol Al2(SO3)3) (2 Al(OH)3)
(294.17¦ g Al2(SO3)3) (1 mol Al2(SO3)3)
= 0.06798¦789 mol Al(OH)3= 0.06799 mol Al(OH)3
c)
Determine the number of grams of Na2SO3 produced
= 12.85¦42 g Na2SO3 = 12.85 g Na2SO3
d) Determine the number of grams of excess reagent left over in the reaction
The method shown below for calculating the amount of NaOH that reacts is slightly different than
what was discussed in class. In this example, we will use the starting amount of Al2(SO3)3 to
calculate the amount of NaOH that reacts. The Al2(SO3)3 is the limiting reactant, so it can also be
used to calculate the amount of NaOH that reacts.
(10.00 g Al2(SO3)3) * (1 mol Al2(SO3)3) *(6 mol NaOH) (39.99¦8 g NaOH) = …
(294.17¦ g Al2(SO3)3) (1 mol Al2(SO3)3)(1 mol NaOH)
= 8.158¦13985 g of NaOH are consumed
EXCESS: 10.00 g NaOH - 8.158¦13985 g of NaOH = 1.84¦186 g NaOH will remain unreacted
= 1.84 g NaOH will remain unreacted
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