Question

How many grams of methanol (CH3OH) would need to be combusted to raise the temperature of 25.0 L of water by 2.25 °C, given the information below? The specific heat of water is 4.18 J/g⋅°C. Assume the density of water is 1.000 g/mL.

2CH3OH(l) + 3O2(g) = 2CO2(g) + 4H2O(l)

∆H° = –1452.8 kJ

Answer 1

we have:

density of water = 1.000 g/mL

volume of water = 25.0 L = 25000 mL

m = density * volume = 25000 g

C = 4.18 J/g.oC

delta T = 2.25 oC

we have below equation to be used:

Q = m*C*delta T

Q = 25000.0*4.18*2.25

Q = 235125 J

Q = 235.125 KJ

This is heat required to raise temperature of water

from reaction,

when 2 mol of CH3OH burns, heat released = 1452.8 KJ

So,

for 235.125 KJ, number of mol of CH3OH = 235.125 * 2 / 1452.8 mol

= 0.3237 mol

Molar mass of CH3OH = 1*MM(C) + 4*MM(H) + 1*MM(O)

= 1*12.01 + 4*1.008 + 1*16.0

= 32.042 g/mol

we have below equation to be used:

mass of CH3OH,

m = number of mol * molar mass

= 0.3237 mol * 32.042 g/mol

= 10.37 g

Answer: 10.37 g