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) (a) For the following precipitation reaction, determine the Limiting Reactant and the species in excess
if 2.99 g of Li2CO3 react with 4.077 g of Ca(C2H3O2)2:
Li2CO3 (aq) + Ca(C2H3O2)2 (aq) ------> 2 LiC2H3O2 (aq) + CaCO3 (s)
Determine the L.R.
2.99 g Li 2CO3 ×
1 mol Li 2CO3 1 mol CaCO3 100.087 g CaCO3
×
×
= 4.05 | 01 g CaCO3
73.89 g Li 2CO3 1 mol Li2CO3
1 mol CaCO3
4.077 g Ca(C2 H 3O 2 ) 2 ×
1 mol Ca(C2 H 3O 2 ) 2
1 mol CaCO 3
100.087 g CaCO 3
×
×
= 2.579 | 90 g CaCO 3
158.167 g Ca(C2 H 3O 2 ) 2 1 mol Ca(C2 H 3O 2 ) 2
1 mol CaCO3
Therefore, Ca(C2H3O2)2 is the limiting reactant and Li2CO3 is in excess.
(b) Now calculate how many grams of ppt are formed:
2.580 g CaCO3
(c ) How many grams of the species in excess are leftover?
First, find the amount of Li2CO3 that reacts.
1 mol Ca(C2 H 3O 2 ) 2
1 mol Li2CO3
73.89 g Li2CO3
4.077 g Ca(C2 H 3O 2 ) 2 ×
×
×
= 1.904 | 63 g Li2CO3
158.167 g Ca(C2 H 3O 2 ) 2 1 mol Ca(C2 H 3O 2 ) 2 1 mol Li2CO3
Mass of leftover Li2CO3 = initial mass – mass used up = 2.99 g – 1.904|63 g
= 1.08|5 g Li2CO3 leftover
= 1.09 g Li2CO3 leftover
7) 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 2521076 g C8H18 in the next calculation, but remember that it will limit us to two significant figures…
(2521076 g C8H18)*(1mol C8H18) (16 mol CO2) (44.010 g CO2)(1 lb ) (1 ton) = 8.6 tons CO2
(114.231 g C8H18)(2 mol C8H18) (1 mol CO2) (453.59g)(2000 lb)
8)
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
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?
(54.52 g CO) * (1 mol CO) * (2 mol H2) (2.016 g H2)
(28.01 g CO) (1 mol CO) (1 mol H2)
= 7.848¦077 g H2 required
EXCESS: 15.21 g H2 – 7.848¦077 g H2 = 7.36¦192 g H2 = 7.36 g H2
(d) Suppose the actual yield is 22.52 g of CH3OH. What is the % yield?
(22.52 /62.36¦80)*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
(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