How Much? The Amount of Chemical Change (DP IB Chemistry)

The balanced equation is:

Mg₃N₂ (s)  + 6H₂O (l) → 3Mg(OH)₂ (aq) + 2NH₃ (aq)

The balanced chemical equation for the thermal decomposition of calcium carbonate is:

CaCO₃ (s)  → CaO(s) + CO₂(g)

The balanced equation is:

2CO (g) + 2NO (g) → 2CO₂ (g) + N₂ (g)

The balanced chemical equation for the formation of hydrogen chloride gas from its component elements is:

H₂ (g) + Cl₂ (g) → 2HCl (g)

False.

When completing reacting masses calculations, the mass unit (e.g. grams, tonnes) does not affect the calculation.

The mass unit only affects the final anwer as it must be given in the appropriate units.

The correct order of steps to complete a reacting mass calculation is:

1. Write a balanced symbol equation.

2. Calculate relevant molar masses.

3. Calculate the moles of the known chemical.

4. Check the molar ratio in the equation.

5. Determine the moles of the target chemical.

6. Calculate the mass of the target chemical.

CuCO₃ CuO + CO₂

The molar ratio of copper carbonate to copper oxide is 1 : 1.

CuCO₃ CuO + CO₂

12.35 g of copper carbonate = 12.35 / 123.5 = 0.1 moles.

The molar ratio of copper carbonate to copper oxide is 1 : 1.

So, 0.1 moles of copper oxide will be formed with a mass of 0.1 x 79.5 = 7.95 g.

2Na + Cl₂ 2NaCl

The molar ratio of chlorine to sodium chloride is 1 : 2.

2Na + Cl₂ 2NaCl

11.7 g of sodium chloride = 11.7 / 58.5 = 0.2 moles.

The molar ratio of sodium to sodium chloride is 1 : 1.

So, 0.2 moles of sodium chloride requires 0.2 moles of sodium with a mass of 0.2 x 23.0 = 4.6 g.

False.

The masses of reactants and products in a reaction can be used to write a balanced equation for the reaction.

How the masses of the reactants and products be used to form a balanced chemical equation:

• Convert all masses to moles

• Find the molar ratio of all chemicals

• Simplify the molar ratio

• Place these values in front of each chemical in the equation

False.

3A + B 2C + 2D

The molar ratio of A : C is not 2 : 3.

There are 3 moles of A and 2 moles of C, which means the ratio is 3 : 2.

The moles, mass, molar mass equation that is needed for reacting mass questions is:

moles = mass / molar mass

6.0 g of magnesium = 6.0 / 24 = 0.25 moles

2Mg + O₂ 2MgO

The ratio of Mg : MgO is 1 : 1

So, 0.25 moles of magnesium oxide can be produced from 6.0 g of magnesium.

M_r (Al₂O₃) = 102.

Moles of aluminium oxide = 51 / 102 = 0.5.

The ratio of Al₂O₃ : Al is 1 : 2.

So, 0.5 moles of aluminium oxide will produce 1.0 moles of aluminium.

So, the maximum mass of aluminum that can be produced is 27 tonnes.

A chemical equation should be balanced before performing reacting mass calculations to ensure that the correct molar ratios are used in the calculations.

The equation for molar gas volume using moles and volume is:

Molar gas volume = volume / moles

The conditions for standard temperature and pressure are:

• 0 ^oC / 273 K

• 100 kPa

The equation to calculate moles using volume and molar gas volume is:

Moles = volume / molar gas volume

At standard temperature and pressure, one mole of gas occupies 22.7 dm³.

The equation connecting moles, volume and molar gas volume is:

Volume = moles x molar gas volume

The amount of space, in dm³, that 1.5 moles of gas occupies is:

• Volume = moles x molar gas volume

• Volume = 1.5 x 22.7 = 34.05 dm³

True.

To convert volume from dm³ to cm³, you multiply by 1000.

A 300 dm³ cylinder contains 300 000 cm³ of gas.

True.

Avogadro’s Law states that equal amounts of gases occupy the same volume of space at the same temperature and pressure.

There are 3 moles of gas in 68.1 dm³.

moles = =3

2.5 moles of gas will occupy 56.75 dm³ at RTP.

• Volume (dm³) = moles x 22.7

• Volume (dm³) = 2.5 x 22.7 = 56.75 dm³

The volume of ammonia is produced from 600 cm³ of hydrogen is 400 cm³.

N₂ (g) + 3H₂ (g)   2NH₃ (g)

• hydrogen : ammonia = 3:2 ratio

• volume of ammonia = 600 cm³ x = 400 cm³

If 250 cm³ of oxygen gas reacted reacted with propane, 150 cm³ of carbon dioxide is formed.

C₃H₈ (g) + 5O₂ (g) → 3CO₂ (g) + 4H₂O (g)

• oxygen : carbon dioxide = 5:3 ratio

• volume of carbon dioxide = 250 cm³ x = 150 cm³

100 cm³ of propane was required to react with 500 cm³ oxygen.

C₃H₈ (g) + 5O₂ (g) → 3CO₂ (g) + 4H₂O (g)

• propane : oxygen = 1:5 ratio

• volume of propane = = 100 cm³

The change in the number of moles of gas is 9 to 4 OR decreases by 5.

4NH₃ (g) + 5O₂ (g) → 4NO (g) + 6H₂O (l)

False.

The molar volume of a gas is the same for all gases at STP (22.7 dm³ mol⁻¹).

Concentration refers to the amount of solid / solute there is in a specific volume of the liquid / solvent.

Using moles, volume in dm³ and concentration, the equation is:

Moles = concentration x volume

To calculate the number of moles of solute present in 2.0 dm³ of a solution whose concentration is 0.15 mol dm^-3:

• Moles = concentration x volume

• Moles = 0.15 x 2.0 = 0.3 moles

The equation for concentration using moles and volume is:

concentration = moles / volume

The concentration of a solution where 1.0 mole of solute is dissolved in 2.0 dm³ of water is:

• Concentration = moles / volume

• Concentration = 1.0 / 2.0 = 0.5 mol dm^-3

The equation to calculate the volume of a solution using moles and concentration is:

Volume = moles / concentration

Volumetric analysis is a process that uses the volume and concentration of one chemical reactant (a standard solution) to determine the concentration of another unknown solution.

Titration is a technique used in volumetric analysis to determine the concentration of an unknown solution using a standard solution of known concentration.

A volumetric pipette is used to measure precise volumes in a titration.

True.

In a titration, a few drops of the indicator are added to the solution in the conical flask before adding the solution from the burette.

Concordant results in titrations are multiple trials that yield very close or identical results, typically within 0.1 cm³.

A standard solution is a solution of accurately known concentration, used as a reference in volumetric analysis.

The formula for calculating concentration in mol dm^-3 is:

Concentration (mol dm^-3) = moles / volume (dm³)

True.

In concentration calculations, you always need to convert cm³ to dm³ by dividing by 1000.

A back titration is a technique used to find the concentration or amount of an unknown substance indirectly by reacting it with an excess of a known reactant and then titrating the excess.

A monoprotic acid is an acid that can donate one proton (H⁺ ion) per molecule in an aqueous solution.

A limiting reactant is the reactant that is completely consumed in a chemical reaction and determines the amount of product formed.

The amount of product formed in a reaction is determined by the limiting reactant.

False.

The limiting reactant is not always the reactant with the smallest mass. It depends on the stoichiometry of the reaction.

A reactant is in excess when it is present in an amount greater than necessary to react with the limiting reactant.

The steps to determine the limiting reactant are:

1. Write the balanced equation for the reaction.

2. Calculate the moles of each reactant.

3. Compare the moles & deduce the limiting reactant.

False.

There can only be one limiting reactant in a reaction.

After the reaction is complete, the reactants in excess remain unreacted.

The presence of a limiting reactant determines the maximum amount of product that can be formed, regardless of the amounts of other reactants present.

True.

The limiting reactant is always completely consumed in a reaction.

An easy way to determine the limiting reactant is:

1. Find the moles of each substance.

2. Divide by the coefficient in the equation.

3. The lowest resulting number is the limiting reactant.

Theoretical yield is the maximum amount of product that can be produced in a chemical reaction, calculated based on the limiting reactant.

Percentage yield compares the actual yield to the theoretical yield.

Actual yield is the recorded amount of product obtained from a chemical reaction.

The actual yield can be determined by experiment only.

Theoretical yield is the amount of product that would be obtained under perfect practical and chemical conditions.

The type of calculation lets you determine theoretical yield is a reacting mass calculation.

The equation for percentage yield is:

(actual yield theoretical yield) x 100

True.

For economic reasons, the objective is to have as high a percentage yield as possible to increase profits and reduce costs and waste.

If the actual yield is 1.50 g and theoretical yield is 2.00g, the percentage yield is:

(1.50 2.0) x 100 = 75%

False.

A percentage yield of 100% is not common in chemical reactions.

Factors that can reduce percentage yield include:

• Incomplete reactions

• Side reactions

• Loss during processing

• Reversible reactions.

A low percentage yield indicates that the reaction is inefficient, with significant loss of product or incomplete conversion of reactants.

Percentage yield can be improved by optimizing reaction conditions, using catalysts, preventing side reactions, and improving separation and purification techniques.

Atom economy is a measure of the efficiency of a chemical reaction in terms of how well reactants are utilised to produce useful products.

Atom economy tells us what percentage of the mass of reactants become useful products.

The atom economy for N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g) is 100% as there is only one product.

True.

The higher the atom economy of a process then the more sustainable that process is

The atom economy for the production of CaO in the reaction is 56%.

The atom economy of a reaction can be increased by selling the by-products of the reaction.

The atom economy for the production of ethene is 100%.

False.

Higher atom economy means less waste production.

Waste reduction in chemical processes is important because it minimises environmental impact and is more sustainable.