Practise Questions

1. Mass of K = 39.1/158.0 × 6.22 = 1.54 gMass of Mn = 54.9/158.0 × 6.22 = 2.16 gMass of O = 16.0/158.0 × 6.22 × 4 = 2.52 g

2. FeCrO4

3. 88.6 g of bromine will be occur together with (100 88.6 = 11.4 g) of phosphorus.Mole ratio of Br : P is found to be 3:1 formula is PBr3.

4. Moles of CaSO3 = Moles of SO2 = 0.124Mass of SO2 = 7.95 g

5. Moles Cl2 = 0.789 Moles NaClO2 = 0.701Limiting Reagent is NaClO2.Moles ClO2 = 0.701 Mass = 47.3 g

6. Moles of formic acid = 3.15/46 = 0.06848Moles O2 = 2.0/22.4 = 0.08929Limiting Reagent is formic acidMoles CO2 = 0.06848 × 2 = 0.1370Volume of CO2 produced = 0.1370 × 22.4 = 3.07dm3

7. a) Moles K2Cr2O7 = 0.0442Conc. = 0.0442/0.250 = 0.177 mol dm3

b) Moles CuSO4.5H2O = 0.240Conc. = 0.240/0.800 = 0.300 mol dm3

8. a) Note that the moles of NaCl remains unchanged moles before dilution = moles after dilution 0.250 × 1.00 = c × 1.0

c = 0.250concentration is 0.250 mol dm–3

b) Total moles of HCl = (0.100 x 0.150) + (0.250 x 0.300) = 0.0900 Total volume = 0.350 dm3

Resultant concentration = 0.0900/0.350 = 0.257 mol dm–3

9. Moles BaSO4 = 0.0215Moles acid req’d = 0.0215 Vol = 35.8 cm3

10. Amount of methanol needed = 0.25 × 5.0 = 1.25 molMass of methanol needed = 32.0 × 1.25 = 40 g0.793 kg dm-3 0.793 g cm-3

Volume of original methanol solution needed = 40 / 0.793 = 50.4 cm3

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TAMPINES JUNIOR COLLEGE

2013 JC1 Chemistry Tutorial

H2 Level Atoms, Molecules and Stoichiometry

Mastery Questions

1. Let the mass of CaSO3 be x g and the mass of MgSO3 be (1.83x) gDecomposition eqn : XSO3 XO + SO2

Moles of CaO formed = Moles of CaSO3 = x/120.2Moles of MgO formed = Moles of MgSO3 = (1.83x)/104.4

Equating masses of the oxide :

Solving, we get x = 1.41 Mass of CaSO3 = 1.41 g or 76.9% Mass of MgSO3 = 0.42 g or 23.1%

2. To form XeF4, Xe 2F2 XeF4

To form XeF6, Xe 3F2 XeF6

Moles of Xe used = 1.77 x 104 Mass of Xe used = 0.0232 gLet the mass of Xe in XeF4 be x g and that of Xe in XeF6 be (0.0232x) g

Equating moles of F2 :

Solving, we have x = 0.0031

Mass of XeF4 = g

Similarly, mass of XeF6 = g

Mass of unreacted Xe = 0.00118 gMass % of XeF4 = (0.00490/0.04368) × 100 = 11.2 %Mass % of XeF6 = (0.0376/0.04368) × 100 = 86.1 %

3 Vol. CO2 formed = 40 cm3

Since 10 cm3 hydrocarbon used, formula is C4Hy.C4Hy + (4 + y/4)O2 4CO2 + y/2 H2O

Initial Gas Volumes / cm3 10 xsFinal Gas Volumes / cm3 0 xs[10(4+y/4)]40

[10 + xs] [xs10(4+y/4) + 40] = 35Solving, we have y = 10

4 Vol. of HCl after mixing = 4.57 dm3

Moles of HCl present = 4.57 x 0.893 = 4.081Moles contributed by 0.315 M HCl = 1.2 x 0.315 = 0.378Moles contributed by HCl of unknown concentration = 4.081 0.378 = 3.703Concentration of this second HCl solution = 3.703/3.37 = 1.10 mol dm–3

5 a) A:B reacts in a 2:1 ratiob) Expt 3: 20 cm3 of A react with 10 cm3 of B. 20 cm3 of B remains.

Expt 5: 20 cm3 of A react with 10 cm3 of B. 20 cm3 of A remains.

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