9 Calculations from chemical equations

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1 Pages Questions 1 molar mass of Fe 2 (SO 4 ) 3 = (2 55.8) + 3 [ (4 16.0)] = g mol 1 molar mass of Fe(OH) 3 = (3 17.0) = g mol 1 amount of Fe 2 (SO 4 ) 3 (in moles) = mass/molar mass = 12.7 g/399.9 g mol 1 = mol moles Fe(OH) 3 = moles Fe 2 (SO 4 ) 3 number of Fe(OH) 3 in equation/number of Fe 2 (SO 4 ) 3 in equation = mol 2 = mol mass of Fe(OH) 3 = moles molar mass = mol g mol 1 = 6.78 g [e] It is a good idea to work out the molar masses of the two substances involved before starting the rest of the calculation. 2 molar mass of AgNO 3 = (3 16.0) = g mol 1 amount (in moles) of AgNO 3 = mass/molar mass = 12.6 g/169.9 g mol 1 = mol ratio of Cu:AgNO 3 = 1:2 so, amount of copper = ½ = mol mass of copper needed = moles molar mass = mol 63.5 g mol 1 = 2.35 g [e] The answer 2.35 g was obtained keeping all the numbers on the calculator during the calculation. If the rounded up value of mol is used, the answer 2.36 g is obtained. Either would score full marks. A common error is to calculate the molar mass of AgNO 3 as g mol 1 because there are 2 moles of it in the equation. This is wrong. The number of moles of a substance depends only on its mass and not on the reaction. The stoichiometry comes into play when moles of one substance are converted to moles of another substance. This type of calculation can be done either by mass ratio or by converting to moles, then using the reaction stoichiometry and finally converting back to mass. The second method fits all types of calculation, so is the better one to use. 3 molar mass of NaNO 3 = (3 16.0) = 85.0 g mol 1 amount of NaNO 3 = mass/molar mass = 33.3 g/85.0 g mol 1 = mol ratio of O 2 :NaNO 3 = 1:2 so, amount of oxygen = ½ = mol volume of oxygen = moles molar volume = mol 25.0 dm 3 mol 1 = 4.90 dm 3 4 ratio of NH 3 :H 2 = 2:3

2 so, theoretical amount of ammonia produced = ⅔ 60 = 40 dm 3 percentage yield = actual yield of product 100/theoretical yield of product percentage yield = /40 = 62.5% [e] When calculating the percentage yield, the yields can be expressed in moles or in mass units. Calculating both yields in moles gives: actual amount (moles) of NH 3 = 2550 g/17.0 g mol 1 = 150 mol so, percentage yield = 150 mol 100/1000mol = 15.0% The percentage yield is not mass of product 100/mass of reactant. 5 amount (in moles) = concentration volume = mol dm /1000 dm 3 = mol [e] Be careful about units. The concentration is in mol dm 3, but the volume is in cm 3. The volume in cm 3 must be divided by 1000 to convert it into a volume in dm 3. 6 volume = moles/concentration = mol/0.106 mol dm 3 = dm 3 = 15.5 cm 3 [e] Volumes of solutions less than 1 dm 3 are usually expressed in cm 3. 7 amount (moles) of H 2 C 2 O 4.2H 2 O = concentration volume = mol dm 3 250/1000 dm 3 = mol molar mass of H 2 C 2 O 4.2H 2 O = (2 12.0) + (4 16.0) + (2 18.0) = g mol 1 mass of hydrated ethanedioic acid needed = moles molar mass = mol g mol 1 = g 8 a) amount (moles) of Fe = mass/molar mass = 4.50 g/55.8 g mol 1 = mol amount of copper(ii) sulfate = concentration volume in dm 3 = 2.00 mol dm dm 3 = mol As is greater than , iron is the limiting reagent. [e] As the substances react in a 1:1 ratio, the one with fewer moles is the limiting reagent. b) ratio of Cu to Fe in the equation = 1:1 so, amount (moles) of copper = amount of iron = mol mass of copper = moles molar mass = mol 63.5 g mol 1 = 5.12 g c) Copper sulfate is in excess, so the solution will still be blue. [e] If copper sulfate had been the limiting reagent, the solution would be colourless at the end of the reaction. 9 a) 2NaOH + H 2 SO 4 Na 2 SO 4 + 2H 2 O [e] The stoichiometric ratio is not 1:1, so the amount of sodium sulfate that would be produced from both sodium hydroxide and sulfuric acid reacting completely has to be calculated. The limiting reagent is the one that gives the smaller amount of product.

3 b) molar mass of sodium hydroxide = = 40.0 g mol 1 amount (moles) of NaOH = mass/molar mass = 21.5 g/40.0 g mol 1 = mol amount (moles) of Na 2 SO 4 that would be produced if NaOH were the limiting reagent = ½ = mol amount (moles) of H 2 SO 4 = concentration volume in dm 3 = 1.00 mol dm dm 3 = mol amount (moles) of Na 2 SO 4 that would be produced if H 2 SO 4 were the limiting reagent = mol, which is more than mol from NaOH Therefore, sodium hydroxide is the limiting reagent. c) molar mass of Na 2 SO 4 = (2 23.0) (4 16.0) = g mol 1 mass of sodium sulfate produced = moles molar mass = mol g mol 1 = 38.2 g d) Sodium hydroxide is the limiting reagent, so there will be an excess of sulfuric acid. Therefore, red litmus will stay red and blue litmus will turn red. [e] Even though there are more moles of sodium hydroxide than sulfuric acid, the sodium hydroxide is the limiting reagent because 2 moles of it are needed for every mole of sulfuric acid. 10 There are dm 3 of CO in 1 dm 3 of air. Assume that the molar volume = 24 dm 3 mol 1. amount of CO = /24 = mol mass of CO = mol 28 g mol 1 = g (= 3.15 μg) 11 mass of water = = 2.48 g moles of water = 2.48 g/18 g mol 1 = mol M r of FeSO 4 = g mol 1 moles of FeSO 4 = 2.98/151.9 = mol = moles of hydrated FeSO 4 ratio moles of water:moles of hydrated FeSO 4 = : = 7.02 number of molecules of water of crystallisation = 7 [e] The answer must be a whole number. 12 a) mass of glucose = g dm = g moles of glucose = g/180 g mol 1 = mol b) number of molecules = mol mol 1 = a) Error 1: It should be dissolved in a smaller amount of water and the solution and washings made up to 250 cm 3 (not dissolved in 250 cm 3 of water). Error 2: A standard flask should have been rather than a beaker. b) moles of Na 2 CO 3.H 2 O in 250 cm 3 = 1.55 g/124 g mol 1 = mol moles of Na 2 CO 3.H 2 O in 25 cm 3 = /250 = mol moles of HCl = = mol

4 [HCl] = mol/ dm 3 = mol dm 3 14 ppm is the mass per million g of solvent. moles of HCl = dm mol dm 3 = mol = mol of KOH mass of KOH = mol 56.1 g mol 1 = g in 100 g mass of KOH in 10 6 g of water = 21.2 g = ppm of KOH Pages Exam practice questions 1 a) volume of CO 2 = dm 3 ( ) moles of CO 2 = dm 3 /24.4 dm 3 mol 1 = = moles CaCO 3 ( ) mass of CaCO 3 = moles molar mass = = g ( ) purity = /1.23 = 96.4% ( ) b) moles CaCO 3 = moles CO 2 = moles of HCl needed = = ( ) volume = moles/concentration = /2 = dm 3 = 11.8 cm 3 ( ) 2 a) A ( ) b) D ( ) [e] The error is ± = ±0.1, so % error = /11.21 = 0.89% c) moles HCl = dm mol dm 3 = mol ( ) d) moles NaOH in 25 cm 3 = mol ( ) moles in excess = = mol ( ) e) initial moles of NaOH = = mol ( ) moles reacted with ibuprofen = = mol ( ) f) moles of ibuprofen = mass of ibuprofen in 25 tablets = mol 228 g mol 1 = 5.54 g ( ) mass in one tablet = 5.54/25 = g = 222 mg ( ) 3 a) i) moles = mass/molar mass ( ) [e] A good way is to check units: g/g mol 1 = mol ii) moles = volume/molar volume ( ) iii) moles = number of particles/avogadro constant ( ) b) D ( ) c) D ( ) d) A ( ) [e] moles Rb = 8.55/85.5 = 0.1; moles O 2 = ( )/32 = 0.1; ratio Rb:O 2 = 1:1 e) B ( )

5 [e] moles CuO = 0.1 = theoretical moles CuSO 4.5H 2 O theoretical mass = g; % = /24.96 = 92% 4 a) Error 1: not rinsing glass rod ( ) some sodium carbonate left on rod ( ) Error 2: not rinsing beaker ( ) some sodium carbonate left in beaker ( ) Error 3: not shaking graduated flask ( ) so variable concentrations of 25 cm 3 samples ( ) b) i) first titre 25.05, second 25.35, third ( ) [e] All must be given to 2 decimal places. ii) moles Na 2 CO 3.10H 2 O = 7.03 g/286 g mol 1 = mol ( ) moles in 25 cm 3 sample = mol ( ) moles HCl = = mol ( ) [HCl] = mol/ dm 3 = mol dm 3 ( )