MC test: Reacting masses & volumes

Multiple choice test on 1.3 Reacting masses & volumes

Use the following 'quiz' to test your knowledge and understanding of this sub-topic. You will need access to a periodic table (Section 6 of the IB data booklet). You may find a calculator useful to answer some of the questions.

If you get an answer wrong, read through the explanation carefully to learn from your mistakes.

What mass of sodium hydroxide will be present in 25.0 cm3 of 2.00 mol dm-3 aqueous sodium hydroxide solution, NaOH(aq)?

M(NaOH) = (22.99 + 16.00 + 1.01) = 40.00 g so 1000 cm3 of 2.00 mol dm-3 NaOH(aq) contains 80.00 g. Hence mass of NaOH in 25.0 cm3 of 2.00 mol dm-3 NaOH(aq) = 80.00 x (25 ÷ 1000) = 2.00 g

 

What volume of 0.250 mol dm-3 potassium hydroxide solution is required to react exactly with 50.0 cm3 of 0.400 mol dm-3 sulphuric acid solution according to the equation below?

H2SO4(aq)  +  2KOH(aq) → K2SO4(aq) + 2H2O(l)

Amount of H2SO4 in 50.0 cm3 of 0.400 mol dm-3 H2SO4 = (50 ÷ 1000) x 0.400 = 0.0200 mol. Since 2 mol of KOH react with 1 mol of H2SO4 amount of KOH required = 0.0400 mol. 1000 cm3 contains 0.250 mol, so volume required = (1000 ÷ 0.250) x 0.0400 = 160 cm3.

 

What will be the resulting concentration in mol dm-3 when 350 cm3 of 0.150 mol dm-3 hydrochloric acid solution is mixed with 450 cm3 of 0.200 mol dm-3 hydrochloric acid solution?

Amount of HCl in 350 cm3 of 0.150 mol dm-3 HCl(aq) = (350 ÷ 1000) x 0.150 = 0.0525 mol. Amount of HCl in 450 cm3 of 0.200 mol dm-3 HCl(aq) = (450 ÷ 1000) x 0.200 = 0.0900 mol. When mixed there will be 0.1425 mol in 800 cm3. Concentration of mixture = (0.1425 ÷ 800) x 1000 = 0.178 mol dm-3.

 

What volume of 1.25 mol dm-3 hydrochloric acid is required to react exactly with excess calcium carbonate to produce 227 cm3 of carbon dioxide gas measured at STP?

1 mol of any gas occupies 22.7 dm3 at STP, so amount of CO2 in 227 cm3 = 1 x 10-2 mol. Since 1 mol of CaCO3 reacts with 2 mol of HCl to form 1 mol of CO2, the amount of HCl required = 2 x 10-2 mol. 1000 cm3 of 1.25 mol dm-3 contains 1.25 mol, so volume of HCl required = (1000 ÷ 1.25) x 2 x 10-2 = 16.0 cm3

 

Paracetamol (acetaminophen) can be synthesised from 4-aminophenol.

In one experiment 7.559g of paracetamol were obtained from 10.913 g of 4-aminophenol. What was the percentage yield?

1 mol of 4-aminophenol produces 1 mol of paracetamol. Amount of 4-aminophenol = 10.913 ÷ 109.13 = 0.100 mol,  so maximum amount of paracetamol that can be obtained from 10.913 g of 4-aminophenol = 0.100 x 151.18 = 15.118 g. Actual yield = 7.559 g so percentage yield = (7.559 ÷ 15.118) x 100 = 50.0%.

 

Which statements explain why real gases do not behave like ideal gases?

I. The volume occupied by the molecules of a real gas is not negligible.

II. There are some attractive forces between the molecules in a real gas.

III. The molecules of real gases collide with the walls of the container.

For ideal gases it is assumed that the volume occupied by the molecules of the gas is negligible and that there are no attractive forces between the molecules. The molecules in both real and ideal gases are assumed to collide with the walls of the container.

 

A sample of argon gas occupies 49.6 cm3 at a temperature of 22 oC and a pressure of 99.3 kPa. What volume (in cm3) will the same amount of argon gas occupy at a temperature of 25 oC and a pressure of 101.2 kPa?

pV = nRT. n and R are constant. At a higher temperature the volume will increase and at a higher pressure the volume will decrease. The temperature must be in Kelvin.

 

What is the maximum volume of hydrogen evolved (measured at STP) when a piece of zinc with a mass of 3.269 g is added to 75.00 cm3 of 1.000 mol dm-3 hydrochloric acid solution? 

I mol of Zn reacts with 2 mol of HCl. Amount of zinc = 3.269 ÷ 65.38 = 0.050 mol. Amount of HCl = (75.0 ÷ 1000) x 1.00 = 0.075 mol so the acid is the limiting reagent. Amount of H2 formed = 0.075 ÷ 2 = 0.0375 mol. 1 mol occupies 22700 cm3 at STP so volume of H2 evolved = 22700 x 0.0375 = 851.3 cm3.

 

Silver chloride is an insoluble salt. What mass of silver chloride will be precipitated when 25.0 cm3 of 0.500 mol dm-3 silver nitrate solution is added to 25.0 cm3 of 0.500 mol dm-3 barium chloride solution?

2AgNO3(aq) + BaCl2(aq) → 2AgCl(s) + Ba(NO3)2(aq)

Amount of Ag+ = (25.0 ÷ 1000) x 0.500 = 0.0125 mol. Amount of Cl− = 2 x (25.0 ÷ 1000) x 0.500 = 0.0250 mol, so Ag+ is the limiting reagent. Amount of AgCl formed = 0.0125 mol. M(AgCl) = (107.87 + 35.45) = 143.32  g mol-1 so mass precipitated = 0.0125 x 143.32 = 1.79 g.

 

When 20.0 cm3 of a hydrocarbon is combusted completely it produces 60.0 cm3 of carbon dioxide and 80.0 cm3 of water vapour. Assuming all the volumes are measured at the same temperature and pressure, what is the molecular formula of the hydrocarbon?

Avogadro's law states that equal volumes of different gases measured at the same temperature and pressure contain the same number of molecules. 1 volume of the hydrocarbon produces 3 volumes of CO2 and 4 volumes of H2O, so it must contain 3 carbon atoms and 8 hydrogen atoms.

 

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