Isotope Problems

1.	Silver atom has two known isotopes existing in nature as 51.82 % 107Ag (atomic mass: 106.9041 u) and 48.18 % 109Ag (atomic mass: 108.9047 u) Calculate the atomic weight of Silver.
	Answer: 107.868 u
2.	There are two known isotopes of Iridium atom: 191Ir with a mass of 190.9609 u and 193Ir which has a mass of 192.9633 u. The atomic weight of Iridium atom is 192.2164. What is the percentage occurrence in nature of 191Ir? What is the percentage occurrence in nature of 193Ir?
	Answer: 37.3% of 191Ir and 62.7% of 193Ir.
3.	Carbon has an atomic weight of 12.011. Its isotopes exist in nature in the following proportions: 98.89% of 12C, 1.11% of 13C. One of its isotope, 12C, has a mass of 12 u. What is the mass of the other isotope?
	Answer: 12.991 u

Here is a new set of chemistry problems that you can use as drill and/or reviewer questions.

Percent Composition Problems

1.	Cd(NO3)2.4H2O
	Answer: H = 2.61% O = 51.87% Cd = 36.44% N = 9.08%
2.	NaHCO3
	Answer: Na = 27.37% H = 1.2% C = 14.3% O = 57.14%
3.	Co(NO3)2
	Answer: Co = 32.21% N = 15.31% O = 52.47%
4.	A laboratory sample was found to contain 68.05 % BaCO3. Compute the amount of C present in 0.67 g of the sample.
	Answer: 0.03 g
5.	A laboratory sample was found to contain 33.56 % KIO3. Compute the amount of I present in 9.91 g of the sample.
	Answer: 1.97 g
6.	A laboratory sample was found to contain 44.72 % PbSO4. Compute the amount of O present in 1.56 g of the sample.
	Answer: 0.15 g
7.	A sample mixture containing ZnCl2 has 4.83 % Zn. Calculate the percentage of ZnCl2 in the sample.
	Answer: 10.07 %
8.	A sample mixture containing CuCO3 has 5.19 % C. Calculate the percentage of CuCO3 in the sample.
	Answer: 53.39 %
9.	Calculate the amount in g of C contained in 5.01 g of CaC4H4O6.
	Answer: 1.28 g of C
10.	Calculate the amount in g of Na contained in 7.43 g of Na2CO3.H2O.
	Answer: 2.75 g of Na

Solute Concentration Problems

1.	Calculate the weight of a 17% by weight CaCl2 solution that will contain 7.61 g CaCl2.
	Answer: 44.76 g solution
2.	Fuming hydrochloric acid has a density of 1.19 g/mL and has 37% HCl by weight. What is the mass of HCl in 55 mL of this acid?
	Answer: 24.22 g HCl
3.	4.69 g of HCl is needed to be obtained from fuming hydrochloric acid (density 1.19 g/mL) which has 37% HCl by weight. Calculate the volume of fuming hydrochloric acid needed.
	Answer: 10.65 mL fuming hydrochloric acid
4.	235 mL of a solution with a concentration of 8 M NaClO4 is needed to be diluted in order to prepare a solution having a concentration of 4 M NaClO4. What must be the final volume of the diluted solution?
	Answer: 470 mL
5.	How many g of ZnCl2 is needed to prepare a 182 mL of a solution containing 3.46 g ZnCl2 per mL?
	Answer: 629.72 g ZnCl2
6.	Calculate the weight of a 8% by weight KI solution that will contain 3.46 g KI.
	Answer: 43.25 g solution
7.	Concentrated sulfuric acid has a density of 1.84 g/mL and has 97% H2SO4 by weight. What is the mass of H2SO4 in 144 mL of this acid?
	Answer: 257.01 g H2SO4
8.	137 mL of a solution with a concentration of 5 M Cu2S is needed to be diluted in order to prepare a solution having a concentration of 1 M Cu2S. What must be the final volume of the diluted solution?
	Answer: 685 mL
9.	4.69 g of HCl is needed to be obtained from fuming hydrochloric acid (density 1.19 g/mL) which has 37% HCl by weight. Calculate the volume of fuming hydrochloric acid needed.
	Answer: 10.65 mL fuming hydrochloric acid
10.	101 mL of a solution with a concentration of 8.24 g K2S2O5 per mL is needed to be diluted in order to prepare a solution having a concentration of 6.45 g K2S2O5 per mL. What must be the final volume of the diluted solution?
	Answer: 129.03 mL

Stoichiometry Problems

1.	Based on the following equation: Cu2S + O2 --> 2Cu + SO2, calculate the number of mol of Cu2S required to yield 4.02 mol of Cu.
	Answer: 2.01 mol of Cu2S
2.	Given the equation: 2KClO3 --> 2KCl + 3O2, how many mol of KClO3 will be needed to produce 7.23 mol of O2?
	Answer: 4.82 mol of KClO3
3.	According to the equation: CaCO3 --> CaO + CO2, calculate the mass of CaCO3 required so that 4.29 mol of CaO will be produced.
	Answer: 429.38 g of CaCO3
4.	Based on the following equation: 2C2H6 + 7O2 --> 4CO2 + 6H2O, how many g of C2H6 will be needed to produce 4.87 mol of H2O?
	Answer: 48.81 g of C2H6
5.	Based on the following equation: CaO + 3C --> CaC2 + CO, how many g of C is required so that 1.2 g of CaC2 is produced?
	Answer: 0.67 g of C
6.	Given the equation: 4KClO3 --> 3KClO4 + KCl, calculate the weight of KClO3 necessary to produce 8.37 g of KClO4.
	Answer: 9.87 g of KClO3
7.	Given the equation: 2KClO3 --> 2KCl + 3O2, calculate the amount of O2 that can be produced from the decomposition of 7.83 g of KClO3.
	Answer: 3.07 g of O2
8.	According to the equation: 4KClO3 --> 3KClO4 + KCl, calculate the weight of KClO3 necessary to produce 2.86 g of KCl.
	Answer: 18.81 g of KClO3
9.	Based on the following equation: 5C + 2SO2 --> CS2 + 4CO, how many g of C will be needed to produce 3.07 mol of CO?
	Answer: 46.09 g of C
10.	Based on the following equation: CaCO3 + 2HCl --> CO2 + CaCl2 + H2O, how many mol of CaCO3 is needed to produce 1.57 g of CO2?
	Answer: 0.04 mol of CaCO3

Here's another set of chemistry reviewer questions.

Stoichiometry Problems

1.	Given the equation: 2C2H6 + 7O2 --> 4CO2 + 6H2O, how many g of C2H6 will be needed to produce 3.13 mol of CO2? Answer: 47.06 g of C2H6
2.	Based on the following equation: 2ZnS + 3O2 --> 2ZnO + 2SO2, how many mol of ZnS is needed to produce 7.95 g of ZnO? Answer: 0.1 mol of ZnS
3.	Given the equation: 2KClO3 + 2KCl --> 3O2 + H2O, how many g of KCl is required so that 3.73 g of H2O is produced? Answer: 30.87 g of KCl
4.	Given the equation: 2KClO3 --> 2KCl + 3O2, calculate the weight of KClO3 necessary to produce 1.57 g of KCl. Answer: 2.58 g of KClO3
5.	Based on the following equation: 3KClO --> 2KCl + KClO3, calculate the amount of KClO3 that can be produced from the decomposition of 7.62 g of KClO. Answer: 3.44 g of KClO3

Percent Composition Problems

1.	Calculate the percentage composition of AlCl3. Solution: Al = 20.24% Cl = 79.76%
2.	Calculate the amount of S contained in 2.88 g of FeSO4.H2O. Solution: 0.54 g of S
3.	Calculate the percentage composition of BaC2O4.2H2O. Solution: H = 1.54% O = 36.73% Ba = 52.54% C = 9.19%
4.	Calculate the amount of Ba contained in 3.76 g of BaI2. Solution: 1.32 g of Ba
5.	Determine the weight of KSCN that will contain 7.69 g of C. Solution: 62.22 g of KSCN
6.	A laboratory sample was found to contain 92.06 % LiNO3. Compute the amount in g of Li present in 2.72 g of the sample. Solution: 0.25 g
7.	A laboratory sample was found to contain 18.83 % PbC2O4. Compute the amount in g of O present in 0.56 g of the sample. Solution: 0.023 g
8.	A sample mixture containing Ba(CH3COO)2 has 1.42 % O. Calculate the percentage of Ba(CH3COO)2 in the sample. Solution: 5.67 %
9.	A sample mixture containing NH4HCO3 has 3.23 % C. Calculate the percentage of NH4HCO3 in the sample. Solution: 21.26 %
10.	Determine the weight of Cd(NO3)2 that will contain 0.84 g of O. Solution: 2.07 g

Limiting Reagent Problems

1.	Based on the following equation: NaCl + AgNO3 --> AgCl + NaNO3, calculate the weight of AgCl which can be produced by the reaction of 7.45 g of NaCl with 7.48 g of AgNO3.Answer: AgNO3 is the limiting reagent; 6.31 g of AgCl can be produced. The weight of excess NaCl is 4.88 g.
2.	Given the balanced equation: 2SO2 + O2 --> 2SO3, calculate the weight of SO3 which can be produced by the reaction of 7.07 g of SO2 with 7.09 g of O2. Answer: SO2 is the limiting reagent; 8.84 g of SO3 can be produced. The weight of excess O2 is 5.32 g.

Formula Calculation Problems

1.	What is the formula weights of the following compounds: (a) KMnO4, (b) KI, (c) CoCl2, (d) Sr(OH)2, (e) (NH4)2S, (f) CdCl2.H2O, (g) B2O3. Answer: (a) 158.0339, (b) 166.0028, (c) 129.8392, (d) 121.6346, (e) 68.1366, (f) 201.3312, (g) 69.6182
2.	Calculate the number of moles of AlCl3 that will contain 6.02 x 1023 atoms of Cl. Answer: 0.33 mol
3.	Calculate the number of moles of PbC2O4 that will contain 6.02 x 1023 atoms of O. Answer: 0.25 mol
4.	How many number of O atoms are contained in 84.94 g of KNO3? Answer: 1.52 x 1024 atoms
5.	How many number of Na atoms are contained in 33.99 g of Na3PO4? Answer: 3.74 x 1023 atoms
6.	Determine the number of molecules of KBrO3 present in 36.77 g of KBrO3. Answer: 1.33 x 1023 molecules
7.	Determine the number of molecules of SbCl3 present in 60.83 g of SbCl3. Answer: 1.61 x 1023 molecules
8.	Calculate the weight in grams of 5.12 mol NaCl. Answer: 299.23 g
9.	Calculate the weight in grams of 0.85 mol CdCl2.H2O. Answer: 171.13 g
10.	Calculate the number of O atoms in 2.44 x 10-4 mol of KCH3COO. Answer: 2.94 x 1020 atoms

Wait...there is more! Tutor Partner has more to offer. Starting today, I'm adding problems on Chemistry here on my blog. The answer is provided for every problem. So stop working on those problems like you're doing homeworks all over again. Take a break; and then take your pick. Stoichiometry Problems 1. Based on the following equation: Na2CO3 + Ba(OH)2 --> BaCO3 + 2NaOH, calculate the number of mol of Na2CO3 required to yield 3.95 mol of NaOH. Answer: 1.98 mol of Na2CO3 2. Based on the following equation: 4FeS2 + 11O2 --> 2Fe2O3 + 8SO2, how many mol of O2 will be needed to produce 7.29 mol of SO2? Answer: 10.02 mol of O2 3. According to the equation: 4Fe + 3O2 --> 2Fe2O3, calculate the mass of O2 required so that 7.16 mol of Fe2O3 will be produced. Answer: 343.67 g of O2 4. According to the equation: 4Fe + 3O2 --> 2Fe2O3, how many g of Fe will be needed to produce 1.75 mol of Fe2O3? Answer: 195.46 g of Fe 5. Given the equation: 4Fe + 3O2 --> 2Fe2O3, how many mol of O2 is needed to produce 8.9 g of Fe2O3? Answer: 0.08 mol of O2 6. For the equation: NaHCO3 + HCl --> NaCl + CO2 + H2O, how many g of NaHCO3 is required so that 1.99 g of NaCl is produced? Answer: 2.86 g of NaHCO3 7. Based on the following equation: CaCO3 --> CaO + CO2, calculate the weight of CaCO3 necessary to produce 7.96 g of CO2. Answer:18.1 g of CaCO3 8. Based on the following equation: 4KClO3 --> 3KClO4 + KCl, calculate the amount of KClO4 that can be produced from the decomposition of 1.68 g of KClO3. Answer: 1.42 g of KClO4 9. According to the equation: 2KClO3 --> 2KCl + 3O2, calculate the number of mol of KClO3 required to yield 5.25 mol of O2. Answer: 3.5 mol of KClO3 10. Based on the following equation: Ni + 4CO --> Ni(CO)4, how many g of Ni will be needed to produce 4.34 mol of Ni(CO)4? Answer: 254.76 g of Ni Percent Composition Problems Calculate the percent composition of the following compounds: 1. PbBr2 Solution: Pb = 56.46% Br = 43.54% 2. FeCl2.4H2O Solution: H = 4.06% O = 32.19% Fe = 28.09% Cl = 35.66% 3. CuSO4 Solution: Cu = 39.81% S = 20.09% O = 40.1% 4. PbC2O4 Solution: Pb = 70.19% C = 8.14% O = 21.68% 5. MgCO3 Solution: Mg = 28.83% C = 14.25% O = 56.93% Limiting Reagent Problems 1. Based on the following equation: Ni + 4CO --> Ni(CO)4, calculate the weight of Ni(CO)4 which can be produced by the reaction of 1.81 g of Ni with 5.94 g of CO. Answer: Ni is the limiting reagent; 5.26 g of Ni(CO)4 can be produced. The weight of excess CO is 2.49 g. 2. Based on the following equation: Cu2S + O2 --> 2Cu + SO2, calculate the weight of Cu which can be produced by the reaction of 4.7 g of Cu2S with 1.83 g of O2. Answer: Cu2S is the limiting reagent; 3.75 g of Cu can be produced. The weight of excess O2 is 0.89 g. 3. Given the balanced equation: CaO + 2HCl --> CaCl2 + H2O, calculate the weight of H2O which can be produced by the reaction of 4.26 g of CaO with 4.8 g of HCl. Answer: HCl is the limiting reagent; 1.19 g of H2O can be produced. The weight of excess CaO is 0.57 g. 4. For the reaction: Ca(HCO3)2 + 2HCl --> CaCl2 + 2CO2 + 2H2O, calculate the weight of H2O which can be produced by the reaction of 9.92 g of Ca(HCO3)2 with 8.44 g of HCl. Answer: Ca(HCO3)2 is the limiting reagent; 2.2 g of H2O can be produced. The weight of excess HCl is 3.98 g. 5. According to the equation: CaO + 2HCl --> CaCl2 + H2O, calculate the weight of CaCl2 which can be produced by the reaction of 3 g of CaO with 6.02 g of HCl. Answer: CaO is the limiting reagent; 5.94 g of CaCl2 can be produced. The weight of excess HCl is 2.12 g.

Analytic Geometry: Distance Between Two Points

Given two points, their distance and the slope of the line joining them are solved.

Problem	Given Points	Distance	Slope
1	P1(2,3) P2(-3,8)	sqrt of 50	1/-1
2	P1(7,1) P2(-5,-7)	sqrt of 208	-2/-3
3	P1(9,4) P2(2,7)	sqrt of 58	3/-7
4	P1(7,-6) P2(-9,-9)	sqrt of 265	-3/-16
5	P1(2,1) P2(9,3)	sqrt of 53	2/7
6	P1(-7,1) P2(-4,-9)	sqrt of 109	-10/3
7	P1(-4,5) P2(2,-4)	sqrt of 117	-3/2
8	P1(4,6) P2(5,-9)	sqrt of 226	-15/1
9	P1(9,8) P2(1,5)	sqrt of 73	-3/-8
10	P1(5,7) P2(7,6)	sqrt of 5	-1/2
11	P1(1,6) P2(2,8)	sqrt of 5	2/1
12	P1(5,1) P2(7,2)	sqrt of 5	1/2
13	P1(-3,2) P2(-6,-8)	sqrt of 109	-10/-3
14	P1(7,9) P2(8,3)	sqrt of 37	-6/1
15	P1(7,8) P2(-9,-4)	sqrt of 400	-3/-4
16	P1(1,8) P2(-7,9)	sqrt of 65	1/-8
17	P1(5,9) P2(6,-7)	sqrt of 257	-16/1
18	P1(2,-7) P2(-7,-5)	sqrt of 85	2/-9
19	P1(-3,9) P2(6,-4)	sqrt of 250	-13/9
20	P1(4,3) P2(7,1)	sqrt of 13	-2/3
21	P1(-5,6) P2(1,-2)	sqrt of 100	-4/3
22	P1(8,3) P2(-5,8)	sqrt of 194	5/-13
23	P1(5,5) P2(-7,-4)	sqrt of 225	-3/-4
24	P1(2,8) P2(3,-6)	sqrt of 197	-14/1
25	P1(6,3) P2(8,8)	sqrt of 29	5/2
26	P1(4,9) P2(7,8)	sqrt of 10	-1/3
27	P1(-7,5) P2(-8,-3)	sqrt of 65	-8/-1
28	P1(6,5) P2(3,8)	sqrt of 18	1/-1
29	P1(9,7) P2(3,3)	sqrt of 52	-2/-3
30	P1(8,-4) P2(-3,-9)	sqrt of 146	-5/-11
31	P1(-8,9) P2(6,-6)	sqrt of 421	-15/14
32	P1(5,9) P2(-1,7)	sqrt of 40	-1/-3
33	P1(9,1) P2(-5,-9)	sqrt of 296	-5/-7
34	P1(2,4) P2(7,1)	sqrt of 34	-3/5
35	P1(8,9) P2(2,3)	sqrt of 72	1/1
36	P1(2,9) P2(7,-2)	sqrt of 146	-11/5
37	P1(4,7) P2(2,4)	sqrt of 13	-3/-2
38	P1(-4,3) P2(-1,-7)	sqrt of 109	-10/3
39	P1(7,5) P2(1,9)	sqrt of 52	2/-3
40	P1(7,-2) P2(-5,-3)	sqrt of 145	-1/-12
41	P1(3,5) P2(2,9)	sqrt of 17	4/-1
42	P1(1,6) P2(4,-9)	sqrt of 234	-5/1
43	P1(4,1) P2(5,5)	sqrt of 17	4/1
44	P1(8,-7) P2(-7,-9)	sqrt of 229	-2/-15
45	P1(1,9) P2(-3,7)	sqrt of 20	-1/-2
46	P1(2,2) P2(-7,-8)	sqrt of 181	-10/-9
47	P1(9,2) P2(5,6)	sqrt of 32	1/-1
48	P1(1,8) P2(7,1)	sqrt of 85	-7/6
49	P1(-9,8) P2(4,-2)	sqrt of 269	-10/13
50	P1(-8,1) P2(-5,-6)	sqrt of 58	-7/3