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## Stoichiometry questions and answers (Page 3)

### Question 1

Copper reacts with concentrated nitric acid as shown by the equation below:

Cu(s) + 4HNO3(aq) → Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)

1. Suggest one observation made when copper reacts with concentrated nitric acid.
2. (i) Suggest the name of the salt of formula Cu(NO3)2.

(ii) Using the equation, determine whether copper is oxidised or reduced when it reacts with concentrated nitric acid.

3. An excess of copper is added to 25.0cm3 of 16.0 mol/dm3 HNO3. Using this information, together with the equation above, calculate the volume of NO2 formed at room temperature and pressure.
4. When heated, Cu(NO3)2 decomposes to form CuO, NO2 and O2. Construct the equation for this reaction.
5. To a sample of Cu(NO3)2(aq), a student adds aqueous ammonia drop by drop until it is in
excess.

(i) Describe what is observed.

(ii) The student repeats the experiment but adds aqueous sodium hydroxide instead of
aqueous ammonia. Describe what is observed.

### Solution 1

1. Either of the following:
• Blue solution is produced (Cu(NO3)2(aq))
• Brown gas is produced (2NO2(g))
• Gas bubbles are produced
• The metal dissolves
2. (i) Copper(II) nitrate.

(ii) To determine whether copper has been oxidised or reduced we first construct an ionic equation for the reaction.

• Cu(s) + 4HNO3(aq) → Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)
• Cu(s) + 4H+ 4NO3(aq) → Cu2+ 2NO3(aq) + 2NO2(g) + 2H2O(l)
• Cu(s) + 4H+ 2NO3(aq) → Cu2+(aq)+ 2NO2(g) + 2H2O(l)

Copper changed from Cu to Cu2+, which means its oxidation state increased from 0 to +2. Therefore Copper has been oxidised.

3. Since the concentration is given in mol/dm3, the units of volume must first be converted to dm3. When converting units from cm3 to dm3 we divide the value by 1000 because 1000 cm3 = 1 dm3.

V = 25/1000 = 0.025 dm3

The next step is to find the number moles of acid that took part in the reaction, using the given data.

• Number of moles of = concentration × volume
• Moles of acid = 0.025 × 16 = 0.4

According to the equation:

Cu(s) + 4HNO3(aq) → Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)

The molar ratio HNO3 : NO2 = 4:2

Using this molar ratio, 0.4 moles of acid produces 0.2 moles of NO2 gas.

∴ Moles of NO2 = 0.2

Then finally we use formula V = Number of moles × molar gas volume

For Cambridge exams, the molar gas volume is 24 dm3, for Zimsec exams its 28dm3.

Volume of NO2 = 0.2 × 28dm3

= 5.6 dm3.

4. 2Cu(NO3)2 → 2CuO + 4NO2 + O2
5. (i) Blue precipitate that turns to dark blue solution in excess.

(ii) Blue precipitate which does not redissolve in excess.

### Question 2

1. Silver nitrate can be prepared by reacting silver oxide with dilute nitric acid.

Ag2O + 2HNO3 → 2AgNO3 + H2O

Excess silver oxide is reacted with 30.0 cm3 of 0.150 mol/dm3 nitric acid. After purification the percentage yield of silver nitrate is 80.0%. Calculate the mass of silver nitrate prepared, giving your answer to three significant figures. [Mr of AgNO3 = 170]

2. Phosphorus can be manufactured from calcium phosphate, Ca3(PO4)2 as follows:

2Ca3(PO4)2 + 6SiO2 + 10C → 6CaSiO3 + 10CO + P4

What is the maximum mass of phosphorus that can be made using 300g of silicon dioxide, SiO2?

### Solution 2

1. From the data given, first convert the volume from cm3 to dm3 and then calculate the number of moles of nitric acid that reacted.

Volume = 30 cm3 ÷ 1000 = 0.03 dm3

Number of moles = concentration × volume

Moles of nitric acid = 0.03 × 0.150 = 0.0045 moles

From the equation:

Ag2O + 2HNO32AgNO3 + H2O

The molar ratio of nitric acid to silver nitrate = 2:2 = 1:1

Since 0.0045 moles of acid reacted, 0.0045 moles of silver nitrate were theoretically produced.

Mass = number of moles × molar mass

The expected mass of silver nitrate produced = 0.0045 × 170

= 0.765 g

Actual mass = % yield × expected mass

= 80% × 0.765

= 0.8 × 0.765

= 0.612g of silver nitrate

2. Number of moles = mass ÷ molar mass

Moles of SiO2 that reacted = 300 ÷ (28 + 16×2)

= 300 ÷ 60

= 5 moles

From the equation:

2Ca3(PO4)2 + 6SiO2 + 10C → 6CaSiO3 + 10CO + P4

The molar ratio of silicon dioxide to phosphorus = 6:1

Using the molar ratio, therefore 5 moles of silicon dioxide reacted to produce 0.83 moles of phosphorus.

Mass = number of moles × molar mass

= 0.83 × (31 × 4)

= 0.83 × 124

= 102.92 g

= 103 g of phosphorus.

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