Calculations Flashcards
The metal lead reacts with warm dilute nitric acid to produce lead(II) nitrate, nitrogen monoxide and water according to the following equation.3Pb(s) + 8HNO3(aq) → 3Pb(NO3)2(aq) + 2NO(g) + 4H2O(I)In an experiment, an 8.14 g sample of lead reacted completely with a 2.00 mol dm–3solution of nitric acid.Calculate the volume, in dm3, of nitric acid required for complete reaction. Give your answer to 3 significant figures.
Mol Pb = 8.14 / 207(.2) (= 0.0393 mol) Mol HNO3 = 0.0393 x 8 / 3 =0.105 mol Vol HNO3 = 0.105 / 2 = 0.0524 (dm3)
In a second experiment, the nitrogen monoxide gas produced in the reaction occupied 638 cm3 at 101 kPa and 298 K.Calculate the amount, in moles, of NO gas produced.(The gas constant R = 8.31 J K–1 mol–1)
101000 (Pa) and 638 x 10-6 (m3)n=pV/RT (=101000x638x10-6 ) ( 8.31x298 )0.026(0) (mol)
Norgessaltpeter was the first nitrogen fertiliser to be manufactured in Norway. It has the formula Ca(NO3)2Norgessaltpeter can be made by the reaction of calcium carbonate with dilute nitric acid as shown by the following equation.CaCO3(s) + 2HNO3(aq) Ca(NO3)2(aq) + CO2(g) + H2O(I)In an experiment, an excess of powdered calcium carbonate was added to 36.2 cm3 of 0.586 mol dm–3 nitric acid.Calculate the amount, in moles, of HNO3 in 36.2 cm3 of 0.586 mol dm–3 nitric acid. Give your answer to 3 significant figures.
0.0212
Calculate the amount, in moles, of CaCO3 that reacted with the nitric acid. Give your answer to 3 significant figures.
0.0106
Calculate the minimum mass of powdered CaCO3 that should be added to react with all of the nitric acid.Give your answer to 3 significant figures.
Mr = 100.1 1.06 g
Norgessaltpeter decomposes on heating as shown by the following equation. 2Ca(NO3)2(s) 2CaO(s) + 4NO2(g) + O2(g)A sample of Norgessaltpeter was decomposed completely.The gases produced occupied a volume of 3.50 × 10–3 m3 at a pressure of 100 kPa and a temperature of 31 °C.(The gas constant R = 8.31 J K–1 mol–1)Calculate the total amount, in moles, of gases produced.
T = 304(K) and P = 100 000 (Pa)100 000 x 3.50 x 10(3) / (8.31 x 304)OR n = PV RT0.139 (mol)
Hence calculate the amount, in moles, of oxygen produced.
0.0276 – 0.0278(mol)
Hydrated calcium nitrate can be represented by the formula Ca(NO3)2.xH2O where x isan integer.A 6.04 g sample of Ca(NO3)2.xH2O contains 1.84 g of water of crystallisation.Use this information to calculate a value for x. Show your working.
4.20 g Ca(NO3)2Ca(NO3)2 H2O 4.20/164 (.1) 1.84/ 18 0.0256 0.102 1 : 3.98 x=4
An unknown metal carbonate reacts with hydrochloric acid according to the following equation.M2CO3(aq) + 2HCl(aq) → 2MCl(aq) + CO2(g) + H2O(l)A 3.44 g sample of M2CO3 was dissolved in distilled water to make 250 cm3 of solution. A 25.0 cm3 portion of this solution required 33.2 cm3 of 0.150 mol dm–3 hydrochloric acid for complete reaction.Calculate the amount, in moles, of HCl in 33.2 cm3 of 0.150 mol dm–3 hydrochloric acid. Give your answer to 3 significant figures.
4.98 x 10-3
Calculate the amount, in moles, of M2CO3 that reacted with this amount of HCl. Give your answer to 3 significant figures.
2.49 x 10-3
Calculate the amount, in moles, of M2CO3 in the 3.44 g sample. Give your answer to 3 significant figures.
2.49 x 10-2
Calculate the relative formula mass, Mr, of M2CO3 Give your answer to 1 decimal place.
138.2
Hence determine the relative atomic mass, Ar, of the metal M and deduce its identity.Ar of M Identity of M
(138 - 60)  2 = 39.1 K/ potassium
In another experiment, 0.658 mol of CO2 was produced. This gas occupied a volume of 0.0220 m3 at a pressure of 100 kPa.Calculate the temperature of this CO2 and state the units.(The gas constant R = 8.31 J K–1 mol–1)
PV = n RT or rearrangedT = 0.022 x 100000 / 0.658 x 8.31402(.3) K (or 129 0C)
In a different experiment, 6.27 g of magnesium carbonate were added to an excess of sulfuric acid. The following reaction occurred.MgCO3 + H2SO4 → MgSO4 + CO2 + H2O Calculate the amount, in moles, of MgCO3 in 6.27 g of magnesium carbonate.
Mr =84.36.27 = 0.074(4) 84.3
Calculate the mass of MgSO4 produced in this reaction assuming a 95% yield.
Mr MgSO4 = 120(.4)Expected mass MgSO4 =0.074(4) x120(.4)= 8.96 g95% yield = (8.96 x 95) / 100 = 8.51 g
Magnesium nitrate decomposes on heating to form magnesium oxide, nitrogen dioxide and oxygen as shown in the following equation.2Mg(NO3)2(s) → 2MgO(s) + 4NO2(g) + O2(g) Thermal decomposition of a sample of magnesium nitrate produced0.741 g of magnesium oxide.Calculate the amount, in moles, of MgO in 0.741 g of magnesium oxide to 3 s.f.
Mr MgO = 40.3 0.741/40.3 = 0.0184
Calculate the total amount, in moles, of gas produced from this sample of magnesium nitrate to 3 s.f.
0.0184 x 5/2 = 0.0460
In another experiment, a different sample of magnesium nitrate decomposed to produce 0.402 mol of gas. Calculate the volume, in dm3, that this gas would occupy at 333K and 1.00 × 105 Pa to 3 s.f.(The gas constant R = 8.31 J K–1 mol–1)
pV=nRTV= (0.402 x 8.31 x 333 ) / 100 0000.0111 11.1 (dm3)
A 0.0152 mol sample of magnesium oxide, produced from the decomposition of magnesium nitrate, was reacted with hydrochloric acid.MgO + 2HCl → MgCl2 + H2OCalculate the amount, in moles, of HCl needed to react completely with the 0.0152 molsample of magnesium oxide to 3 s.f.
0.0152 x 2 = 0.0304
This 0.0152 mol sample of magnesium oxide required 32.4 cm3 of hydrochloric acid for complete reaction. Use this information and your answer to part (c) (i) to calculate the concentration, in mol dm–3, of the hydrochloric acid.
0.938 mol dm-3
There are several oxides of nitrogen.An oxide of nitrogen contains 25.9% by mass of nitrogen. Determine the empirical formula of this oxide.
O=74.1% 25.9 / 14 74.1 / 16 1.85 4.63 1 2.5N2O5
Define the term relative atomic mass.An organic fertiliser was analysed using a mass spectrometer. The spectrum showedthat the nitrogen in the fertiliser was made up of 95.12% 14N and 4.88% 15NCalculate the relative atomic mass of the nitrogen found in this organic fertiliser. Give your answer to two decimal places.
Average/mean mass of (1) atom(s) (of an element) realtive to 1/12 mass of one atom of 12C((95.12 x 14) + (4.88 x 15)) / 100= 14.05
Ammonium sulfate reacts with sodium hydroxide to form ammonia, sodium sulfate and water as shown in the equation below.(NH4)2SO4(s) + 2NaOH(aq) → 2NH3(g) + Na2SO4(aq) + 2H2O(l)A 3.14 g sample of ammonium sulfate reacted completely with 39.30 cm3 of a sodium hydroxide solution.Calculate the amount, in moles, of (NH4)2SO4 in 3.14 g of ammonium sulfate.
Mr = 132.1 0.0238
Hence calculate the amount, in moles, of sodium hydroxide which reacted.
0.0476
Calculate the concentration, in mol dm–3, of the sodium hydroxide solution used.
1.21
Calculate the percentage atom economy for the production of ammonia in the reaction between ammonium sulfate and sodium hydroxide.
34 x 100 / 212.1= 16.0(3)%
Ammonia is manufactured by the Haber Process. N2 + 3H2  2NH3Calculate the percentage atom economy for the production of ammonia in this process.
100%
A sample of ammonia gas occupied a volume of 1.53 × 10–2 m3 at 37 °C and a pressure of 100 kPa.(The gas constant R = 8.31 J K–1 mol–1)Calculate the amount, in moles, of ammonia in this sample.
PV =nRT or n= PV / RTn = (100000 x 1.53 x 10-2 ) / 8.31 x 310= 0.59(4)
Glauber’s salt is a form of hydrated sodium sulfate that contains 44.1% by mass of sodium sulfate. Hydrated sodium sulfate can be represented by the formula Na2SO4.xH2O where x is an integer. Calculate the value of x.
(Na2SO4) (44.1% ) H2O 55.9%44.1/142.1 55.9/8 0.310 3.11 =1 =10x =10
When aqueous aluminium sulphate reacts with aqueous barium chloride, a white precipitate of barium sulphate is formed. An equation for this reaction is shown below.Al2(SO4)3(aq) + 3BaCl2(aq) → 3BaSO4(s) + 2AlCl3(aq)Hydrated aluminium sulphate has the formula Al2(SO4)3.xH2O, where xH2O represents thewater of crystallisation.A sample of hydrated aluminium sulphate of mass 20.0 g was dissolved in water and the solution made up to 250 cm3.An excess of aqueous barium chloride was added to a 25.0 cm3 portion of this aluminium sulphate solution.All the sulphate ions reacted to form a precipitate of barium sulphate.When filtered, washed and dried, the mass of the barium sulphate precipitate was 2.10 g.Calculate the number of moles of barium sulphate (Mr = 233.4) in the precipitateCalculate the number of moles of aluminium sulphate in the 25.0 cm3 portion of the solution and in the original 20.0 g sample.Molesin25.0cm3 ……………………………………………………………………………………….. ………………………………………………………………………………………………………………….. Moles in original sample
Moles BaSO4 = 2.10 ÷ 233.4 = 9.00 × 10-3Moles Al2(SO4)3 in 25 cm3 = 3.00 × 10-3; Moles Al2(SO4)3 in sample = 3.00 × 10-2;
Calculate the Mr of this sample of hydrated aluminium sulphate and hence deduce the value of x in Al2(SO4)3.xH2O(If you have been unable to obtain an answer for the number of moles of aluminium sulphate in the original sample, in part (a) (ii), you may assume that the answer is 3.17 × 10–2 mol. This is not the correct value.)
Mr of Al2(SO4)3⋅xH2O = 20.0 ÷ 3.00 × 10-2; = 666.9Mr of Al2(SO4)3 = 342(.3);Mr of xH2O = 666.9 – 342.3 = 324.6X = 18 / Al2(SO4)3.18H2O
A 1.37 g sample of a barium compound, X, contains 0.835 g of barium and 0.146 g of carbon, the rest being oxygen.Use these data to calculate the empirical formula of X
Mass O = 0.389 Ba C O0.835/137.3 1.46/12 0.389/160.00608 0.0122 4.00 1 2 4BaC2O4
A 1.04 g sample of methanol vapour, CH3OH (Mr = 32.0), was maintained at a pressure of 101 kPa and a temperature of 350 K. Use the ideal gas equation to calculate the volume,in cm3, that this sample of methanol vapour would occupy under these conditions.(The gas constant R = 8.31 J K–1 mol–1)
pV = nRTMoles methanol = n = 1.04 / 32 (= 0.0325 mol)V = nRT/p = 0.0325 × 8.31 × 350 /101000= 9.36 × 10-4 m3= 936 cm3;
