Practice Problems: Set #3-Solutions

Practice Problems: Set #3-Solutions IIa) Balance the following equations:(10) 1) Zn (s) + H 3 PO 4 (aq) Zn 3 (PO 4 ) 2 (s) + H 2 (g) 3Zn (s) + 2H 3 PO 4 (aq) Zn 3 (PO 4 ) 2 (s) + 3H 2 (g) 2. Mg 3 N 2 (s) + H 2 O (l) NH 3 (g) + Mg(OH) 2 (s) Mg 3 N 2 (s) + 6H 2 O(l) 2NH3(g) + 3Mg(OH) 2 (s) 3. KMnO 4 (aq) + HCl(aq) Cl 2 (g) + MnO 2 (s) + KCl(aq) + H 2 O(l) 4H 2 O(l) 2 KMnO 4 (aq) + 8HCl(aq) 3Cl 2 (g) + 2 MnO 2 (s) + 2 KCl(aq) + 4. NH 3 (g) + O 2 (g) NO(g) + H 2 O (l) 4NH 3 (g) +5O 2 (g) 4NO(g) + 6H 2 O (l) 5. Al 2 S 3 (s) + H 2 O (l) Al(OH) 3 + H 2 S(g) Al 2 S 3 (s) + 6H 2 O (l) 42Al(OH) 3 + 3H 2 S(g) IIb) Write a balanced chemical equation for each of the following:(8) 1. Solid copper reacts with solid sulfur to form solid copper(i) sulfide. 2Cu(s) + S(s) Cu 2 S (s) 2. Solid magnesium reacts with aqueous copper (I) nitrate to form aqueous magnesium nitrate and solid copper. Mg(s) + 2Cu (NO 3 )aq Mg(NO3) 2 aq + 2Cu(s) 3. Gaseous dinitrogen pentoxide decomposes to form nitrogen dioxide and oxygen gas. 2N 2 O 5 (g) 4NO2 + O 2 4. Aqueous hydrochloric acid reacts with solid manganese (IV) oxide to form aqueous manganese (II) chloride, liquid water, and chlorine gas.

4HCl(aq) + MnO 2 MnCl2 (aq) + 2H 2 O + Cl 2 (g) IIIa) Write the chemical formula for each of the following(5) 1.Potassium Dichromate K 2 Cr 2 O 7 2. ammonium phosphate (NH 4 ) 3 PO 4 3. Phosphoric acid H 3 PO 4 4. Baking Soda NaHCO 3 5. Iron (III) sulfide Fe 2 S 3 Multiple Choice: 1. The balanced equation 2Cu(s) + O 2 (g) 2CuO(s) tells us that 1 mol of Cu A. reacts with 1 mol of O 2 B. produces one mol of CuO C. must react with 32g of O 2 D. cannot react with oxygen E. produces 2 mol of CuO 2. A 3.0 mol sample of KClO 3 was decomposed according to the equation 2KClO 3 (s) 2KCl(s) + 3O 2 (g) How many moles of O 2 are formed assuming 100% yield? A. 2.0 mol

B. 2.5 mol C. 3.0 mol D. 4.0 mol E. 4.5 mol 3. What mass of carbon dioxide would be produced when 10.0 g of butane reacts with an excess of oxygen in the following reaction? 2C 4 H 10 (g) + 13 O 2 (g) 8CO 2 (g) + 10 H 2 O(g) A. 7.57g CO 2 B. 30.3g CO 2 C. 40.0g CO 2 D. 352g CO 2 E. none of these 4. How many molecules of carbon dioxide would be formed if 6.75g of propane is burned in the following reaction? C 3 H 8 (g) + 5O 2 (g) 3 CO 2 (g) + 4H 2 O (g) A. 5.54 x 10 23 molecules B. 1.39 x 10 23 molecules C. 20.3 x 10 23 molecules D. 2.77 x 10 23 molecules E. 3.89 x 10 23 molecules 5. Consider the following reaction: 2A + B 3C + D 3.0 mol A and 2.0 mol B react to form 4.0 mol C. What is the percent yield of the reaction? A. 50% B. 67% C. 75% D. 89% E. 100% 6. When NH 3 is prepared from 28g N 2 and excess H 2, the theoretical yield of NH 3 is 34g. When this reaction is carried out in a given experiment, only 30g is produced. What is the percentage yield? A. 6% B. 12% C. 14% D. 82% E. 88%

7. The empirical formula for the compound having the formula H 2 C 2 O 4 is A. COH B. COH 2 C. C 2 H 2 D. C 2 O 4 H 2 E. CO 2 H 8. Which of the following has the largest percent by mass of carbon? A. CaCO 3 B. CO 2 C. CH 4 D. NaHCO 3 9. A compound is analyzed and found to contain 12.1% carbon, 16.2% oxygen, and 71.7% chlorine (by mass). Calculate the empirical formula of this compound. A. COCl B. COCl 2 C. CO 2 Cl D. CO 2 Cl 2 E. COCl 4 10. Calculate the molecular formula of a compound with the empirical formula CH 2 O and a molar mass of 150g/mol. A. C 2 H 4 O 2 B. C 3 H 6 O 3 C. C 4 H 8 O 4 D. C 5 H 10 O 5 E. C 6 H 12 O 6 11. A certain compound is found to have the percent composition (by mass) of 85.63% C and 14.37% H. The molar mass of the compound was found to be 42.0 g/mol. Calculate the empirical and the molecular formulas. A. C 2 H 3 and C 4 H 6 B. CH and C 3 H 3 C. CH 2 and C 3 H 6 D. CH 3 and C 2 H 6 E. C 2 H 6 and C 3 H 9

12. The correct formula for arsenic(iii) oxide is A. As 3 O 3 B. As 2 O 3 C. As 3 O 2 D. AsO E. As 3 O 13. Identify the incorrect statement among the following: A. A mole is that quantity of a substance that contains 6.02x10 23 particles of that substance. B. One mole of any substance contains the same number of particles as one mole of any other substance. C. One mole of any substance contains the same number of particles as the number of atoms in exactly 12 grams of carbon-12. D. A mole is a quantity of a substance that has a mass of exactly 12 grams. E. All of the above are correct. 14. In the sum of 54.34 + 45.66, the number of significant figures is A. 2 B. 3 C. 4 D. 5 (100.00- trailing zeros significant, when number is written with a decimal point) E. 6 15. The number 0.005802 expressed in scientific notation is A. 5.82 x 10 3 B. 5.802 x 10 3 C. 5.82 x 10-3 D. 5.802 x 10-3 E. 5802 x 10-6 16. Which of the following is a binary compound? A. O 2 B. HCN C. H 2 SO 4 D. H 2 S

E. NaOH 17. Select the correct statement about 43 Ti A. The nucleus contains 22 protons and 26 neutrons, mass number =48 B. The nucleus contains 22 protons and 21 neutrons, atomic number =22 C. The nucleus contains 21 protons and 22 neutrons, mass number =43. D. The nucleus contains 22 protons and 43 neutrons, atomic number 22 E. None of the above are correct. 18. The sum of the coefficients when the following equation is balanced is A. 4 A. 7 B. 8 C. 9 D. 11 BaSO 4 + K 3 PO 4 Β a 3 (PO 4 ) 2 + K 2 SO 4 19. The name for HBrO (aq) is A. hydrogen bromous acid B. bromous acid C. bromic acid D. hypobromous acid E. perbromic acid 20. The reaction AgNO 3 (aq) + NaCl (aq) AgCl(s) + NaNO 3 (aq) is a(n) ------------------ reaction. A. precipitation B. acid-base C. oxidation-reduction D. none of these E. single replacement

21. A reaction that involves a transfer of electrons is called a(n) ------- reaction A. precipitation B. acid base C. oxidation-reduction D. double displacement E. none of these IIb) Write a balanced chemical equation for the following: (4) 1. Aqueous hydrochloric acid reacts with solid manganese (IV) oxide to form aqueous manganese (II) chloride, liquid water, and chlorine gas. MnO 2 (s) + 4HCl(aq) MnCl2 (aq) + 2H 2 O(l) + Cl 2 (g) 2. Liquid nitric acid decomposes to reddish brown nitrogen dioxide gas, liquid water, and oxygen gas. (This is why bottles of nitric acidbecome yellow upon standing). 4HNO 3 (l) 4NO2(g) + 2H 2 O (l) + O 2 (g) IIIa) Balance the following chemical equations.(6) Na 2 O 2 (s) + H 2 O(l) NaOH(aq) + O 2 (g) 2Na 2 O 2 (s) + 2H 2 O(l) Mg 3 N 2 (s) + H 2 O (l) NH 3 (g) + Mg(OH) 2 (s) 4NaOH (aq) + O2 (g) Mg 3 N 2 (s) + 6H 2 O(l) 2NH3(g) + 3Mg(OH) 2 (s) FeCl 3 (aq) + KOH (aq) Fe (OH) 3 + KCl (aq) FeCl 3 (aq) + 3KOH(s) Fe(OH) 3(s) + 3KCl (aq) IIIb) Give the name and calculate the molar mass for each of the following:(4) 1.Cr 2 O 3 Chromium (III) oxide; 152.00g 2. NaClO : Sodium Hypochlorite- 74.44g

IV)Solve the following: 1.Calculate the theoretical yield of NO and the percentage yield of NO if 49.2g of NO 2 yield 8.90 grams of NO in the following reaction:(3 points) 3NO 2 (g) + H 2 O (g) 2HNO 3 (g) + NO(g) gno 2 mol NO2 mol NO gno Factors: 46.0g NO 2 /molno 2 ; 1molNO/3mol NO 2 ; 30.0g NO/molNO So: 49.2gNO 2 x1molno 2 X 1mol NO X 30.0gNO = 10.7g NO-theoretical yield 46.0g NO 2 3mol NO 2 1mol NO Therefore: Actual yield/theoretical yield X 100 = Percent yield or X 100 = 83.2% yield. 8.9 g/10.7g 2. Nitrogen gas can be prepared by passing gaseous ammonia over solid copper (II) oxide at high temperatures. The products of the reaction are solid copper and water vapour. How many grams of nitrogen are formed when 18.1g of NH 3 are reacted with with 90.4g of CuO? (5 points) Step1: 2NH 3 (g) + 3CuO(s) N 2 (g) + 3Cu(s) + 3H 2 O(g) Step2: From masses of reactants available compute moles of NH 3 (molar mass = 17.03g) and CuO (molar mass=79.55g) Therefore: 18.1g NH 3 x 1mol NH 3 = 1.06 mol NH 3 and 90.4 gcuo x 1mol CuO = 1.14 mol CuO 17.03gNH 3 79.55 g CuO Step 3: To determine which reactant is limiting, we use mole ratio between CuO and NH 3 So: if 2mol NH 3 corresponds to 3CuO Then 1.06 mol NH 3? i.e. 1.06 mol NH 3 x 3 mol CuO = 1.59 mol CuO 2 mol NH 3 How much CuO do we have? How much we need? Moles of CuO available = 1.14; this is < moles of CuO needed (1.59)

Therefore, CuO is the limiting reactant and is used in calculating yield of N 2 formed. Step 4: 1.1mol CuO x 1molN 2 = 0.380 mol N 2 3mol CuO Step 5: 0.380mol N 2 x 28.02g N 2 (molar mass) = 10.6g N 2 1mol N 2 BONUS QUESTION (6 points) Finish the following table. (Use the periodic chart given to you) Nuclear Symbol # of Protons # of Neutrons # of electrons Atomic # Mass # 34 S 16 18 16 16 34 51 V 23 28 23 23 51 27 Al 13 14 13 13 27