Moles and reactions

At the end of this section, you should be able to deduce the quantities of

reactants and products from balanced equations

Key information Stoichiometry is the molar relationship between the relative quantities of substances taking part in a reaction.

Balanced equation gives information about:* the reacting quantities that are needed to prepare a required quantity of a product* the quantity of products formed by reacting together of known quantities of reactants.

In dealing with reacting quantities, the following steps should be considered 1. Write the balanced reaction equation if not given 2. Work out the moles of the known reacting quantities3. Work out the mass or any other quantity required

RecallThere are three ways to work out an amount in moles

1. From mass: n = m or m = n x M

M

2. From gas volumes:

n = V(dm3) or V(dm3) = n x 24.024.0

3. From solutions:

n = c x V(dm3)

Examples (reacting masses) 1. 8.00g of aluminium reacted with excess oxygen to

produce aluminium oxide. a) write a balanced equation b) find the number of moles of aluminium oxide formed.c) what was the mass of aluminium oxide produced?

Answer

a) 4Al + 3O2 2Al2O3 4mol 3 mol 2 mol 2 mol 1.5 mol 1 mol

b) Moles of aluminium, n = m M

n = 8.00 = 0.2963 27

From the reaction, 2 mols of Al produces 1 mol of Al2O3

Therefore 0.2963 mols of Al will produce = 0.2963. 2 = 0.1482mols Al2O3.

c) mass = n x M

M(Al2O3) = 27 x 2 + 16 x3 = 102

mass of Al2O3 produced = 102 x 0.1482 = 15.12g

2. What mass of copper oxide is obtained by heating 4.94g of copper carbonate?CuCO3(s) CuO(s) + CO2(g)( hint. Find the moles, mole ratio of quantities and the mass)

Answerfrom n = m ÷ MM(CuCO3) =64 + 12 + 16 x 3 = 124

n = 4.94 ÷ 124 = 0.0398 mol of CuCO3

From the reaction, 1 mol of CuCO3 produces 1 mol of CuOTherefore 0.0398 mol of CuCO3 produces 0.0398 mol of CuOmass of CuO = n x M

M(CuO) = 64 + 16 = 80

= 0.0398 x 80 = 3.18g of CuO

• 3. Sodium and chorine react to form sodium chloride: 2Na (s)+ Cl 2(g) 2NaCl(s) (a) Calculate the amount, in mol, in 2.925 g NaCl.

n = m/M = 2.925 = 2.925 = 0.0500 mol 23.0 + 35.5 58.5

(b) Calculate the amounts, in mol, of Na and Cl2 that would form 2.925 g NaCl.

equation: 2Na (s)+ Cl2(g) 2NaCl(s) amounts in equation 2 mol : 1 mol : 2 mol Amounts required : 0.050mol 0.025mol 0.050mol

(c) Calculate the masses of Na and Cl2 that would form 2.925 g NaCl.

Na : mass = n x M = 0.050 x 23 = 1.15 g

Cl2 : mass = n x M = 0.025 x (35.5+35.5) = 0.025 + 71 = 1.775 g

(4) Zinc carbonate decomposes to produces carbon dioxide and 4.2 g zinc oxide.

(a) write a balanced reaction equation

(b) find the moles of zinc carbonate that decomposed

(c) what mass of zinc carbonate produced the 4.20 g of zinc oxide.

Answers(a) Equation: ZnCO3 ZnO + CO2

(b) n = m MM(ZnO) = 65.4 + 16.0 = 81.4

mole of ZnO, n = 4.20 = 0.0516 mol 81.4

reaction equation: ZnCO3 ZnO + CO2

amounts in equation: 1 mol: 1 mol: 1 molamounts required: 0.0516 0.0516 0.0516therefore amount of zinc carbonate that decomposed was 0.0516 mol(c)mass = n x M M(ZnCO3) = 65.4 + 12.0 + (16.0x3) = 125.4mass of zinc carbonate = 125.4 x 0.0516 = 6.47 gtherefore 6.47 g zinc carbonate is required

Reacting masses and gas volumesIf a substance is a solid, mass is used to measure the amount needed for a reaction. However, if a reactant or product is a gas, it unlikely it could be weighed; it would be easier to measure its volume.

According to Avogadro’s Law: Equal volumes of gases measured at the same temperature and pressure contain the same number of molecules.

This is extremely useful when relating the volume of gas to an equation.

ExampleWhat volume of oxygen reacts exactly with 100cm3 of hydrogen at the same temperature and pressure?

Answerthe equation for the reaction is: 2H2(g) + O2(g) 2H2O(l)the mole ratio 2 mol: 1 mol: 2 mol

volume required 100cm3 50cm3 100cm3

The equation indicates that half the number of oxygen molecules as hydrogen molecules is needed. Therefore, 50cm3 of oxygen reacts with 100cm3 of hydrogen

2. What volume of oxygen reacts exactly with 30 cm3 of methane and what volume of carbon dioxide is obtained at the same temperature and pressure?

AnswerThe equation for the reaction is:CH4 (g) + 2O2 (g) CO2(g) + 2H2O(l)

the mole ratio is 1 mol: 2 mol: 1 mol: 2 mol

volume of gas

30cm3 60cm3 30cm3 60cm3

therefore 60 cm3 of oxygen reacted with the 30 cm3 of methane to produce 30 cm3 of carbon dioxide

Molar volumeIf a volume of gas contains a fixed number of molecules, then it is useful to know what volume contains the Avogadro number of molecules. This volume is known as Molar volume. Molar volume depends on standard temperature of 0oC (273K) and a pressure of 101.3 kPa.Under the above conditions, molar volume is taken as 24000 cm3 or (24 dm3)mol-1.

Main statementAt room temperature and pressure ( r.t.p), 1 mol of a gas occupies 24000 cm3 or 24dm3

(Hint: you need to find the number of moles of the gas in order to calculate the molar volume at r.p.t, and where necessary use mole ratio from the balanced reaction equation)

EXAMPLES1. 2.50g of calcium carbonate was heated and decomposed as in the equation below.CaCO3(s) CaO(s) + CO2(g)(a) calculate the amount, in mol, of CaCO3 that decomposed.

n = m = 2.50 = 2.50 = 0.025 mol M {40 + 12 + (3x16)} 100

(b) calculate the amount, in mol, of carbon dioxide formed.Reaction equation: CaCO3(s) CaO(s) + CO2(g)mole ratio: 1 mol: 1 mol: 1 molamounts of mole required 0.025 0.025 0.025

i.e. 0.025 mol of carbon dioxide formed.

(c) calculate the volume of carbon dioxide that formed.V = n x 24.0dm3 = 0.025 x 24 = 0.60 dm3

2. Calculate the volume of hydrogen produced when excess dilute sulphuric acid is added to 5.00 g of zinc. Assume that under the conditions of the reaction 1 mol of gas has a volume of 24 dm3.

Answer:amount in moles of zinc used, n = m = 5.00 = 0.0765 mol M 65.4the equation is: Zn(s) + H2SO4(aq) ZnSO4(aq) + H2(g)mole ratio 1 mol 1 mol 1 mol 1 molamounts required 0.0765 mol 0.0765 mol

it follows that 0.0765 mol of hydrogen is produced

molar volume = 24.0 dm3

volume of hydrogen produced, V = n x 24.0 = 0.0765 x 24 = 1.836dm3

3. Calculate the volume of hydrogen produced when excess dilute sulphuric acid is added to 5.00 g of aluminium. Under the conditions of the reaction, 1 mol of gas has a volume of 24.0 dm3.

Answeramount of mole of Al used, n = m = 5.00 = 0.185 mol M 27.0the equation for the reaction is: 2Al(s) + 3H2SO4(aq) Al2(SO4)3(aq) + 3H2(g) mole ratio 2 mol 3 mol 1 mol 3 mol simplest ratio 1 mol 1.5 mol 0.5 mol 1.5mol

the equation indicates that 1 mol of aluminium produces 1.5 mol of hydrogen.

Amount in moles of hydrogen produced = 1.5 x 0.185 = 0.278 mol

hence, volume of hydrogen produced, V = n x 24.0 = 0.278 x 24 = 6.667 dm3

4. When reacted with excess dilute hydrochloric acid, what mass of calcium carbonate produces 1 dm3 of carbon dioxide? Assume that 1 mol of of gas has a volume of 24 dm3

Answeramount in moles of carbon dioxide = 1 = 0.0417 mol 24the equation for the reaction is: CaCO3(s) + 2HCl(aq) 2CaCl(aq) + CO2(g) + H2O(l)

mole ratio: 1 mol 2 mol 2 mol 1 mol 1 molfrom the reaction equation 1 mol of carbon dioxide is produced by the reaction of 1 mol of calcium carbonate with excess hydrochloric acid.Therefore amount in moles of CaCO3 used = 0.0417 molmass of CaCO3 used, m = n x MM(CaCO3) = 40. 0 + 12.0 + (3 x 16.0) = 100hence, mass of CaCO3 used = n x M = 0.0417 x 100 = 4.17 g

5. A gas syring has a maximum capacity of 100 cm3. Calculate the mass of copper(II) carbonate that would have to be heated to produce enough carbon dioxide to just fill the syringe at r.t.pAnswersyringe volume of 100 cm3 if equivalent to 100 = 0.00417 mol 24000thus, volume of carbon dioxide to fill the syringe is 0.00417 molthe equation for the reaction is:

CuCO3(s) CuO(s) + CO2(g)mole ratio 1 mol 1 mol 1 mol

amounts required 0.00417 mol 0.00417 molthus, 0.00417mol of carbon dioxide is obtained by heating 0.00417mol of copper carbonate. Mass of copper carbonate, m = n x M = 0.00417 x ( 63.5 + 12 + 16x3) = 0.00417 x 123.5 = 0.515 g

6. Calculate the volume of 8 g of methane at r.t.p

Answer

amount in mol of methane, CH4, n = m = 8 = 8 M 12 + (4 x 1) 16

amount in mol of methane = 0.5 mol

volume of 1 mol of methane at r.t.p = 24.0 dm3

hence, volume of 0.5 mol of methane = 0.5 x 24 = 12 dm3

Reacting mass, gas and solution volumes

You will need to recall some of the formulae used earlier on and use them in such calculations. Rearrangement of the formulae may be necessary in some instances in order to do other calculations.

Recall

1. From mass: n = m or m = n x M

M

2. From gas volumes:

n = V(dm3) or V(dm3) = n x 24.024.0

3. From solutions:

n = c x V(dm3) or c = n V(dm3)

1. A chemist reacted 0.23g of sodium with water to form 250 cm3 of aqueous sodium hydroxide. Hydrogen gas was also produced. The equation is shown below. 2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g)(a) calculate the amount, in mol, of Na that reacted. n = m = 0.23 = 0.010 mol M 23 (b) Calculate the volume of H2 formed at room temperature and pressure.The equation for the reaction is 2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g)mole ratio 2 mol 1 mol amounts required 0.010mol 0.005mol

thus, 0.010 mol of Na produces 0.005 mol H2 gas.Volume of hydrogen gas formed at r.t.p, V = n x 24000 cm3

= 0.005 x 24000 = 120 cm 3

(c ) Calculate the concentration, in mol/dm3, of NaOH(aq) formed.Reaction equation: 2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g)mole ration: 2 mol 2 molamount required 0.010 mol 0.010molthus, 0.010 mol Na formed 0.010 mol of NaOH solutionconcentration of NaOH, in mol/dm3, c = n = 0.010 mol = 0.04 mol/dm3

V dm3 0.250 dm3

2. Balance the equation below.MgCO3(s) + HNO3(aq)(aq) Mg(NO3)2(aq) + H2O(l) + CO2(g)(b) 2.529 g of MgCO3 reacts with an excess of HNO3.(i) what volume of CO2, measured at RTP, is formed?(ii) The final volume of the solution is 50 cm3. What is the concentration, in mol/dm3, of Mg(NO3)2 formed

Now try questions 1 – 3

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