# Can someone help me solve this question? (1 Viewer)

#### spyder101

##### New Member
Students were asked to predict the final mass of the methanol left inside the spirit burner if 300ml
of water were to be heated by 15.0 ̊C. If the initial mass of methanol was measured as 20.0g and the
molar heat of combustion is -726KJ/Mol, what would the students answer with?
Could you provide working out that would be really appreciated.

#### jazz519

##### Moderator
Moderator
This is a molar heat of combustion question. There are two formulas:
delta H = mCdeltaT

delta Hc = delta H / n(fuel)

In your question you are given:
m = 300 g = 0.3 kg
C = 4.18 J/g/K (specific heat capacity of water)
delta T = 15 ̊C

sub into first formula:
delta H = (0.3)(4.18)(15) = 18.81 kJ

The question also states: delta Hc = 726 kJ/mol

Sub these into 2nd formula:
726 = 18.81 / n(fuel)
n(fuel) = 18.81 / 726 = 0.0259... moles

Fuel here is methanol, so n(methanol) = 0.0259... moles
Find the mass: m(CH3OH) = n x MM = (0.0259...)(12.01+4x1.008+16) = 0.83017... g

That's mass of fuel used. So remaining mass in spirit burner = 20 - 0.83017... = 19.2 g (3sf)

• Drdusk

#### spyder101

##### New Member
This is a molar heat of combustion question. There are two formulas:
delta H = mCdeltaT

delta Hc = delta H / n(fuel)

In your question you are given:
m = 300 g = 0.3 kg
C = 4.18 J/g/K (specific heat capacity of water)
delta T = 15 ̊C

sub into first formula:
delta H = (0.3)(4.18)(15) = 18.81 kJ

The question also states: delta Hc = 726 kJ/mol

Sub these into 2nd formula:
726 = 18.81 / n(fuel)
n(fuel) = 18.81 / 726 = 0.0259... moles

Fuel here is methanol, so n(methanol) = 0.0259... moles
Find the mass: m(CH3OH) = n x MM = (0.0259...)(12.01+4x1.008+16) = 0.83017... g

That's mass of fuel used. So remaining mass in spirit burner = 20 - 0.83017... = 19.2 g (3sf)
makes complete sense now thanks