What amount of heat is required to vaporize 143.45 g of ethanol (C₂H₅OH)? (∆Hvap = 43.3 kJ/mol)

135 KJ.

Explanation:

The following data were obtained from the question:

Mass of ethanol, C₂H₅OH = 143.45 g

Heat of vaporisation, ∆Hvap = 43.3 kJ/mol

Heat required (Q) =.?

Next, we shall determine the number of mole in 143.45 g of ethanol (C₂H₅OH). This is can be obtained as follow:

Mass of ethanol, C₂H₅OH = 143.45 g

Molar mass of ethanol, C₂H₅OH = (12×2) + (1×5) + 16 + 1

= 24 + 5 + 16 + 1

= 46 g/mol

Mole of ethanol, C₂H₅OH =?

Mole = mass /Molar mass

Mole of ethanol, C₂H₅OH = 143.45/46

Mole of ethanol, C₂H₅OH = 3.118 moles

Finally, we shall determine the heat required to vaporize the ethanol, C₂H₅OH as follow:

Heat of vaporisation, ∆Hvap = 43.3 kJ/mol

Mole of ethanol, C₂H₅OH (n) = 3.118 moles

Heat required (Q) =.?

Q = n•∆Hvap

Q = 3.118 × 43.3

Q = 135 KJ

Therefore, the heat required is 135 KJ

- lead + phosphoric acid = hydrogen + lead(ii) phosphate