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Revision Topic 1 chemistry

Revision Topic 1 chemistry

Assessment

Presentation

•

Chemistry

•

11th Grade

•

Practice Problem

•

Hard

Created by

Duman Bakayev

Used 1+ times

FREE Resource

8 Slides • 15 Questions

1

Revision DP2 chemistry
media

Topic 1,4,9,10

2

Topic 1 - stoichiometry

You need to use various formulae but all of them are around mole.

A mole is a concept used in chemistry that is equal to 6.02 *1023 of particles.

Formulae you need to know:

n=N/NA, n=m/M, n=V/Vm, n=C*V, yield=mactual/mtheoretical,

mass%=(msubstance/mtotal)*100

PV=nRT - is given in the booklet

3

How do you solve a problem?

1 Realise what you were given

2 Calculate what can be calculated

3 Find a way to the value that is needed to be calculated (usually you have 1 or 2 step questions only, the multi-step processes are divided into smaller questions)

4

The question asks to calculate molar concentration which is C=n/V

V is given 500 cm3 . n is missing. we also have 0.200 g the mass of lithium. We know that there is n=m/M formula. It is the only thing we can do with the mass, hence n(Li)=0.200/6.94=0.0288 mol

Now that we have the mole we can use the equation and see that the mole of Li is equal to the mole of LiOH (same ratio) meaning that n(Li)=n(LiOH)

C(LiOH)=0.0288/0.500=0.0576 mol dm-3 is the answer


Example

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5

Question 1

5.00 g of an impure sample of hydrated ethanedioic acid, (COOH)2•2H2O, was dissolved in water to make 1.00 dm3 of solution. 25.0 cm3 samples of this solution were titrated against a 0.100moldm-3 solution of sodium hydroxide using a suitable indicator.

(COOH)2 (aq) + 2NaOH (aq) → (COONa)2 (aq) + 2H2O (l)

6

Fill in the Blanks

Calculate the amount, in mol, of NaOH in 14.0cm3 of 0.100moldm-3 solution.

Type answer...

7

Fill in the Blanks

Calculate the amount, in mol, of ethanedioic acid in each 25.0 cm3 sample.

Type answer...

8

Fill in the Blanks

Determine the percentage purity of the hydrated ethanedioic acid sample.

Type answer...

9

Question 2

2.478 g of white phosphorus was used to make phosphine according to the equation:

P4(s)+3OH−(aq)+3H2O(l)→PH3(g)+3H2PO2−(aq)

10

Fill in the Blanks

2.478 g of white phosphorus was used to make phosphine according to the equation:

P4(s)+3OH−(aq)+3H2O(l)→PH3(g)+3H2PO2−(aq)

Calculate the amount, in mol, of white phosphorus used.

Type answer...

11

Fill in the Blanks

2.478 g of white phosphorus was used to make phosphine according to the equation:

P4(s)+3OH−(aq)+3H2O(l)→PH3(g)+3H2PO2−(aq)

This phosphorus was reacted with 100.0cm3 of 5.00mol dm−3 aqueous sodium

hydroxide. Deduce, showing your working, which was the limiting reagent.

Type answer...

12

Fill in the Blanks

2.478 g of white phosphorus was used to make phosphine according to the equation:

P4(s)+3OH−(aq)+3H2O(l)→PH3(g)+3H2PO2−(aq)

This phosphorus was reacted with 100.0cm3 of 5.00mol dm−3 aqueous sodium

hydroxide. Deduce, showing your working, which was the limiting reagent.

Determine the excess amount, in mol, of the other reagent.

Type answer...

13

Fill in the Blanks

2.478 g of white phosphorus was used to make phosphine according to the equation:

P4(s)+3OH−(aq)+3H2O(l)→PH3(g)+3H2PO2−(aq)

Determine the volume of phosphine, measured in cm3 at standard temperature and pressure, that was produced.

Type answer...

14

Fill in the Blanks

Question 3

The oxide formed in the reaction with air contains 43.6 % phosphorus by mass. Determine the empirical formula of the oxide, showing your method.

Type answer...

15

Fill in the Blanks

Question 4

Calculate the number of hydrogen atoms in 1.00 g of propan-2-ol.

Type answer...

16

Question 5

3.26 g of iron powder are added to 80.0 cm3 of 0.200 mol dm−3 copper(II) sulfate solution. The following reaction occurs:

Fe (s) + CuSO4 (aq) → FeSO4 (aq) + Cu (s)

17

Fill in the Blanks

Determine the limiting reactant showing your working.

Type answer...

18

Fill in the Blanks

The mass of copper obtained experimentally was 0.872 g. Calculate the percentage yield of copper.

Type answer...

19

Fill in the Blanks



Type answer...

20

Question 6

When heated in air, magnesium ribbon reacts with oxygen to form magnesium oxide.

The reaction was carried out in a crucible with a lid and the following data was recorded:

Mass of crucible and lid = 47.372 ±0.001 g

Mass of crucible, lid and magnesium ribbon before heating = 53.726 ±0.001 g

Mass of crucible, lid and product after heating = 56.941 ±0.001 g

21

Fill in the Blanks

Calculate the amount of magnesium, in mol, that was used.

Type answer...

22

Fill in the Blanks

Determine the percentage uncertainty of the mass of product after heating.

Type answer...

23

Fill in the Blanks

Assume the reaction is the only one occurring and it goes to completion, but some product has been lost from the crucible. Deduce the percentage yield of magnesium oxide in the crucible.

Type answer...

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Revision DP2 chemistry
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Topic 1,4,9,10

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