Define limiting reagent .

A limiting reagent is the reactant that is completely consumed in a reaction. It determines the maximum amount of product that can be formed and is not present in excess.

What is the molar volume of any gas at room temperature and pressure (RTP), and what are the RTP conditions?

At RTP (20 °C, 1 atm), one mole of any gas occupies 24.0 dm 3 . This is Avogadro's hypothesis applied to gases.

True or False? Percentage yield compares the actual yield with the maximum theoretical yield.

True. Percentage yield = (actual yield / theoretical yield) x 100. It shows how much of the expected product was actually obtained.

Concentration (mol dm -3 ) = .......... divided by .......... . To convert cm 3 to dm 3 , divide by .......... .

Concentration (mol dm -3 ) = moles of solute divided by volume of solution (dm 3 ) . To convert cm 3 to dm 3 , divide by 1000 .

Define percentage yield .

Percentage yield is the ratio of the actual yield obtained experimentally to the predicted theoretical yield, expressed as a percentage. It is calculated as: (actual yield / theoretical yield) x 100.

25.0 cm 3 of 0.050 mol dm -3 Na 2 CO 3 is neutralised by 20.0 cm 3 HCl. What is the concentration of HCl? (Na 2 CO 3 + 2HCl → 2NaCl + H 2 O + CO 2 )

Moles Na 2 CO 3 = 0.025 x 0.050 = 0.00125 mol. Moles HCl = 2 x 0.00125 = 0.00250 mol. Concentration HCl = 0.00250/0.020 = 0.125 mol dm -3 .

True or False? Equal volumes of different gases at the same temperature and pressure contain the same number of molecules.

True. This is Avogadro's hypothesis. It means that mole ratios from balanced equations apply directly to gas volumes at the same conditions.

The number of moles of a substance = .......... divided by .......... . A mass of 6.0 g of Mg (M r = 24.3) gives .......... moles.

The number of moles of a substance = mass (g) divided by molar mass (g mol -1 ) . A mass of 6.0 g of Mg (M r = 24.3) gives 0.247 moles.

In a reaction of 9.2 g Na with 8.0 g S to form Na 2 S, which is the limiting reagent? (2Na + S → Na 2 S; A r : Na = 23.0, S = 32.1)

Moles Na = 9.2/23.0 = 0.40 mol. Moles S = 8.0/32.1 = 0.25 mol. To react 0.40 mol Na requires only 0.20 mol S. S is in excess, so Na is the limiting reagent .

A LevelChemistryCambridge (CIE)Flashcards2. Atoms, Molecules & StoichiometryReacting Masses & Volumes (of Solutions & Gases)

Reacting Masses & Volumes (of Solutions & Gases) (Cambridge (CIE) A Level Chemistry): Flashcards

Exam code: 9701

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Define limiting reagent.

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Cards in this collection (9)

Define limiting reagent.
A limiting reagent is the reactant that is completely consumed in a reaction. It determines the maximum amount of product that can be formed and is not present in excess.
What is the molar volume of any gas at room temperature and pressure (RTP), and what are the RTP conditions?
At RTP (20 °C, 1 atm), one mole of any gas occupies 24.0 dm³. This is Avogadro's hypothesis applied to gases.
True or False?
Percentage yield compares the actual yield with the maximum theoretical yield.
True.
Percentage yield = (actual yield / theoretical yield) x 100. It shows how much of the expected product was actually obtained.
Concentration (mol dm^-3) = .......... divided by .......... . To convert cm³ to dm³, divide by .......... .
Concentration (mol dm^-3) = moles of solute divided by volume of solution (dm³). To convert cm³ to dm³, divide by 1000.
Define percentage yield.
Percentage yield is the ratio of the actual yield obtained experimentally to the predicted theoretical yield, expressed as a percentage. It is calculated as: (actual yield / theoretical yield) x 100.
25.0 cm³ of 0.050 mol dm^-3 Na₂CO₃ is neutralised by 20.0 cm³ HCl. What is the concentration of HCl? (Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂)
Moles Na₂CO₃ = 0.025 x 0.050 = 0.00125 mol. Moles HCl = 2 x 0.00125 = 0.00250 mol. Concentration HCl = 0.00250/0.020 = 0.125 mol dm^-3.
True or False?
Equal volumes of different gases at the same temperature and pressure contain the same number of molecules.
True.
This is Avogadro's hypothesis. It means that mole ratios from balanced equations apply directly to gas volumes at the same conditions.
The number of moles of a substance = .......... divided by .......... . A mass of 6.0 g of Mg (M_r = 24.3) gives .......... moles.
The number of moles of a substance = mass (g) divided by molar mass (g mol^-1). A mass of 6.0 g of Mg (M_r = 24.3) gives 0.247 moles.
In a reaction of 9.2 g Na with 8.0 g S to form Na₂S, which is the limiting reagent? (2Na + S → Na₂S; A_r: Na = 23.0, S = 32.1)
Moles Na = 9.2/23.0 = 0.40 mol. Moles S = 8.0/32.1 = 0.25 mol. To react 0.40 mol Na requires only 0.20 mol S. S is in excess, so Na is the limiting reagent.

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