Problem Set III Stoichiometry - Solutions

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1 Chem 121 Problem set III Solutions - 1 Problem Set III Stoichiometry - Solutions molecular mass of ethane = 2(12.011) + 6(1.008) = g 4. molecular mass of aniline = 6(12.011) + 7(1.008) = g/mol 5. Let the molar mass of X be x. x = 31.1 g/mol and the element is P. 6. Let atomic mass of M be x solving for x = 137 g/mol which is Ba how would you check your answer? 7. Let atomic mass of M be x 4 M + 3 O 2 2 M 2 O 3 solving for x = g/mol which is Nd how would you check your answer? 8. Molecular mass of C 2 H 5 OH is (2x x ) = Molecular mass of Ca(HCO 3 ) 2 is ( x x x15.999) = %Ca = 24.72%, %C = 14.82%, %H = 1.24%, %O = 59.22%

2 Chem 121 Problem set III Solutions - 2 Molecular mass of MgNH 4 PO 4 is ( x x15.999) = Mg = 17.70%, N = 10.20%, H = 2.936%, P = 22.56%, O = 46.61% 9. - use CO 2 to get mass of C and H 2 O to get mass of H quick check to see that C and H are the only components of this compound: g g = g OK if this added up to < g, we would know that there must a 3 rd element present, for which we could get the mass by difference 10. The first step is to find the percentage composition of the unknown, as follows: moles CO 2 moles C in CO 2 moles C in compound mass C in compound percent C percent C = moles H 2 O moles H in H 2 O moles H in compound mass H in compound percent H percent H = Percentage of O in unknown = The next step is to find the empirical formula from the percentage composition. I will assume I have g of the unknown for convenience. Element Relative mass Relative number of moles (atoms) Divide by the smallest number C 40.0 H 6.72 O 53.3 The empirical formula is C x H y O z = C 1 H 2 O 1 or CH 2 O To find the molecular formula, the molecular mass needs to be known. Let's say that the molecular mass has been measured by mass spectrometry and found to be amu. Since the empirical formula is CH 2 O, the empirical mass is x = amu. The ratio of molecular mass to empirical mass is So the molecular formula is (CH 2 O) 6 or C 6 H 12 O 6.

3 Chem 121 Problem set III Solutions Take a g sample Atom Mass (g) Moles Divide by smallest Best ratio Na / = Al / = F / = Na 3 AlF Element Relative mass Relative number of moles (atoms) Divide by the smallest number La O The empirical formula is La x O y = La 2 O you don t have to come up with an equation but it can help Ti + S Ti x S y a) mass of product = g g = g product 100% of Ti used resulted in product formation the % of Ti in the product is: and all the rest is S = %S b. for the empirical formula, require moles sulphur was in excess, use the % S in the product to get the amount of sulphur that must have reacted: or could have got g S from g g = g S and Ti 0.184/0.184 S 0.367/0.184 Ti 1 S 1.99 ~ TiS 2

4 14. Mo 2 O 3 Mo x O y 12.64g 13.48g The increase in mass is oxygen, so mass of extra O is = 0.84 g Chem 121 Problem set III Solutions - 4 so total mass of oxygen in new oxide is 2.529g g = 3.369g so mass of Mo in Mo x O y = 13.48g g = g Element Mass Relative number of moles (atoms) Divide by the smallest number Whole number ratio Mo O Empirical formula of the oxide is MoO N 2 O 5 + H 2 O 2 HNO 3 Mg 2 C H 2 O 2 Mg(OH) 2 + C 3 H 4 PCl H 2 O H 3 PO HCl 16 Cr(s) + 3 S 8 (s) 8 Cr 2 S 3 (s) Au 2 S 3 (s) + 3 H 2 (g) 2 Au(s) + 3 H 2 S(g) 6 NH 4 ClO 4 (s) + 10 Al(s) 3 N 2 (g) + 6 HCl(g) + 9 H 2 O(g) first write what you know from the question (you should know that bromine molecules, in fact all the halogens, exist as diatomic molecules) Na + Br 2 NaBr - then balance it 2Na + Br 2 2NaBr we know that when you combust a hydrocarbon, we get CO 2 and H 2 O and the oxidant in combustion is always oxygen (almost always): C 4 H 10 + O 2 CO 2 + H 2 O - now balance it: C 4 H / 2 O 2 4CO 2 + 5H 2 O - but we don t want fractional coefficients: 2C 4 H O 2 8CO H 2 O st we write what we know in an equation: NH 3 + CuO N 2 + Cu - 2 nd we balance it but wait, we ve only got H and O on the reacting side there must be the formation of water in this reaction can balance it now: 2NH 3 + 3CuO N 2 + 3Cu + 3H 2 O - can now do a c

5 Chem 121 Problem set III Solutions CHECK: The sum of the masses of reactants = the sum of the masses of products Reactants: 79.3 g g = g Products: g g = g NH 4 ClO Al 5 Al 2 O N HCl + 9 H 2 O C 57 H 110 O O CO H 2 O 22. Let mass BaO 2 = (x) g, thus mass BaCO 3 = ( x)g Mass BaO formed: NH CuO N Cu + 3 H 2 O Mass nitrogen gas formed: CuO is limiting, NH 3 is in excess and 10.6g of nitrogen is formed Mass NH 3 remaining = 18.1 g g = 5.2 g

6 Chem 121 Problem set III Solutions XeF H 2 O 2 Xe + 4 HF + O g 50.0g mass of HF formed thus XeF 2 is limiting and H 2 O is in excess mass HF formed is 0.236g mass of H 2 O remaining = 50.0g g = 49.9g NH 3 (g) + 5 O 2 (g) 4 NO(g) + 6 H 2 O(l) 2.00g 4.50g mass of NO formed 26. This problem can be solved in a couiple of ways, but the easiest is if it is treated as a empirical formula. Percentage of MgSO 4 in hydrate = = 48.8 So take 100g of hydrate molecule Mass Moles Divide by Closest whole number ratio (g) smallest MgSO H 2 O and so there are 7 molecules of water for every molecule of MgSO 4 in the hydrate. limiting reagent is AgNO 3, KCl is in excess and 4.2g AgCl is formed. mass KCl remaining = 5.0g g = 2.8g thus O 2 is limiting and NH 3 is in excess mass NO formed is 3.38g mass of NH 3 remaining = 2.00g g = 0.08g 27. Percentage Yield = 1.21g * 100 / 1.25 g = 96.8% st write the equation: N 2 + H 2 NH 3 - then balance it: N 2 + 3H 2 2NH 3 - then determine limiting reactant:

7 Chem 121 Problem set III Solutions - 7 with this easy an equation, can see that for 4 mol N 2 would need 3 x 4 = 12 mol H 2 H 2 is limiting - then calculate the theoretical yield on the basis of 6.0 mol H 2 : 29. a) C 6 H 6 + Br 2 C 6 H 5 Br + HBr 30.0g 65.0g mass of C 6 H 5 Br formed thus C 6 H 6 is limiting and Br 2 is in excess mass C 6 H 5 Br formed is 60.3g b) mass of Br 2 remaining = 65.0g g = 3.6g st we write out the equation: - 2 nd we balance it: H 2 O + KO 2 KOH + O 2 2H 2 O + 4KO 2 4KOH + 3O 2 - then we determine which reactant is limiting: i) assume H 2 O is limiting ii) assume KO 2 is limiting: so KO 2 is limiting since it forms the least amount of product and the theoretical yield = 19.7 g KOH - 3 rd calculate the % yield: we re told how much iron is formed in the 2 nd reaction we need to calculate the amount of CO required to get this amount of Fe, using the molar relationships from the 2 nd equation then we can calculate how much carbon is required to generate that much CO from the first equation

8 Chem 121 Problem set III Solutions M 1 V 1 = M 2 V we know M 1 and V 1 and M 2 we solve the dilution eqn for V 2 and subtract the initial volume of HCl soln from final volume to determine the amount of water to add: water added = 100 ml soln required 5.00 ml 12.0 M HCl = 95 ml water added st calculate the # of moles of AgNO 3 present then find the # of moles of K 2 CrO 4 to react with it finally, use molarity to find the volume of interest 40.

9 Chem 121 Problem set III Solutions determine the # mol KMnO 4, then mol Na 2 C 2 O 4 required for the reaction then use definition of percentage 42. H 2 SO NaOH Na 2 SO H 2 O convert mass KHP to moles KHP which converts to mole NaOH and M NaOH soln 44. Step 1. Balance the equations 2Cu 2 O 4Cu + O 2 2CuO 2Cu + O 2 Step 2. Define the variables for CuO and Cu 2 O. Let the amount of Cu 2 O be X g and so the amount of CuO is ( X) g. Step 3: Mass of Cu formed from: So the total mass of Cu formed is X g (1.000-X) g = g Step 4: Solve for X X = g = mass of Cu 2 O Therefore percent Cu 2 O in mixture is 45. Let mass of Mg = x, so mass of Zn = ( x) g Mg + 1/2 O 2 MgO Zn + 1/2 O 2 ZnO

10 Chem 121 Problem set III Solutions - 10 Percentage of Zn in the mixture is 60.3%, and percent Mg is 39.7% start with an unbalanced equation to get an idea of what s going on: KCl + MgCl 2 + H 2 SO 4 HCl - we can write a balanced equation for the titration: HCl + NaOH H 2 O + NaCl we know that mol NaOH = mol HCl = mol Cl - we also know that: g KCl + g MgCl 2 = g so let x = g KCl, then g MgCl 2 must be (0.502 x) then we can calculate the mol of Cl that each compound contributes - the 2 compounds added together donated all the Cl in the HCl formed: x (0.502 x) = g Cl x x = x = and x = = g KCl and g MgCl 2 = = g MgCl 2

Mass Relationships in Chemical Reactions

Mass Relationships in Chemical Reactions Chapter 3 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Micro World atoms & molecules Macro World grams Atomic mass