5 2 quantitative analysis

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:

The 2 common units for concentration are:

To convert g dm-3 to mol dm-3, you divide the concentration in g dm-3 by the molar mass.

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

False. To convert dm3 into cm3, you multiply by 1000.

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

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

To calculate the number of moles present in 2.5 dm3 of a solution which has a concentration of 0.30 mol dm-3:

The balanced equation is: H2SO4 + 2KOH → K2SO4 + 2H2O

27.50 cm3 is 0.0275 dm3.

The ratio of acid to alkali is 1:3. H3PO4 + 3NaOH → Na2PO4 + 3H2O

The concentration in mol/dm3 of a solution of HCl with a volume of 23.55 cm3 and contains 0.00375 moles is: Volume in dm3 = 0.02355 dm3 Concentration = = 0.16 mol/dm3

The four steps in a titration calculation are:

Using moles, volume in dm3 the equation is: Concentration =

0.03905 dm3 is 39.05 cm3.

0.05 moles of HCl would be required as this reaction is a 1:1 ratio. HCl + NaOH → NaCl + H2O

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%

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.

To calculate atom economy: Atom economy =

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.

False. Reactions with low atom economies are unsustainable and not economically attractive as they use up too much raw material and produce a lot of waste.

Three factors companies consider to improve efficiency are:

Percentage yield is the amount of product actually obtained from a reaction compared to the theoretical maximum amount, expressed as a percentage.

False. Fast reaction rates are desirable attributes in industrial chemical processes.

Companies can improve the atom economy of a process by selling or reusing the waste products, or by considering alternative production methods with more useful by-products.

A reversible reaction is a chemical reaction that can proceed in both forward and backward directions, potentially reaching a state of equilibrium.

True. Altering reaction conditions can shift the equilibrium position of a reversible reaction.

Waste disposal is expensive as it requires chemicals, equipment, space and transport.

A high temperature would favour the formation of products if the forward reaction is endothermic.

The equation for molar gas volume using moles and volume is: Molar gas volume = volume / moles

The conditions for room temperature and pressure are:

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

At rtp, the number of moles of gas present in 48 dm3 is:

One mole of any gas at room temperature and pressure occupies:

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:

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 2 moles of gas in 48 dm3. moles = = 2

2.5 moles of gas will occupy 60 dm3 at RTP. volume (dm3) = moles x 24 volume (dm3) = 2.5 x 24 = 60 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)