Getting the most from reactants test questions - Higher Chemistry Revision

1

What are feedstocks?

Raw materials or chemicals used to produce more profitable chemicals

Products from chemical reactions that are sold for profits

Catalysts used to make processes more efficient

2

Which of the following would be grounds for reconsidering a chemical process that has been designed?

Producing a side-product

Producing toxic gases

Producing waste that can decompose or biodegrade naturally

3

\[CuO + H_{2} \rightarrow Cu + H_{2}O\]

What mass of copper (formula mass = 63.5 g) would be produced when 16 g of copper(II) oxide (formula mass = 79.5 g) reacts with excess hydrogen gas?

315.5 g

12.77 g

6.39 g

4

Which of the following factors will affect the molar volume of a gas?

Temperature

Formula mass of the gas

Type of intermolecular bonding present

5

The ester propyl ethanoate (formula mass = 102 g) can be formed from propanol (formula mass = 60 g) and ethanoic acid (formula mass = 60 g) as shown below:

\[C_{3}H_{7}OH + CH_{3}COOH\rightleftharpoons C_{3}H_{7}OOCCH_{3} + H_{2}O\]

If 3 g of propanol react to form 1.8 g of propyl ethanoate, what is the percentage yield for this process?

35 per cent

45 per cent

55 per cent

6

The ester propyl ethanoate (formula mass = 102 g) can be formed from propanol (formula mass = 60 g) and ethanoic acid (formula mass = 60 g) as shown below:

\[C_{3}H_{7}OH + CH_{3}COOH\rightleftharpoons C_{3}H_{7}OOCCH_{3} + H_{2}O\]

Why is the percentage yield never 100 per cent for this process?

The ester destroys the catalyst

Some of the product is lost

The reaction is reversible

7

The ester propyl ethanoate (formula mass = 102 g) can be formed from propanol (formula mass = 60 g) and ethanoic acid (formula mass = 60 g) as shown below:

\[C_{3}H_{7}OH + CH_{3}COOH\rightleftharpoons C_{3}H_{7}OOCCH_{3} + H_{2}O\]

What is the atom economy for this process?

70 per cent

85 per cent

100 per cent

8

Calculate the mass of nitrogen dioxide (formula mass = 46 g) which is produced when 119 cm³ of nitrogen (formula mass = 28 g) is sparked in excess oxygen (assume molar volume is 23.8 l mol^-1).

0.46 g

4.6 g

460 g

9

Calculate which reactant is in excess when 0.655 g of zinc (formula mass = 65.5 g) reacts with 20 cm³ of hydrochloric acid, concentration 0.5 mol l^-1.

\[Zn+2HCl \rightarrow ZnCl_{2} + H_{2}\]

Zinc

Hydrochloric acid

Neither

10

Calculate which reactant is in excess when 25 cm³ of 1 mol l^-1 sulfuric acid is mixed with 25 cm³ of 1 mol l^-1 sodium hydroxide.

\[H_{2}SO_{4}+ 2NaOH \rightarrow Na_{2}SO_{4} + 2H_{2}O\]

Sodium hydroxide

Sulfuric acid

Neither