Help much appreciated (1 Viewer)

[Jimmy2064]

This is question 17 from the 2011 paper, and I'm stuck as to how to go about it. Any help would be greatly appreciated.


17 The molar heat of combustion of pentan-1-ol is 2800 kJ mol−1
. A quantity of pentan-1-ol
was combusted, generating 108 kJ of heat.
What mass of pentan-1-ol was combusted?
(A) 2.29 g
(B) 2.86 g
(C) 3.32 g
(D) 3.40 g

 

[planino]

D)

Use equivalent fractions i.e.

2800kJ/1mol = 108kJ/ x mol

therefore (rearranging), x mol = 27/700

m(pentan-1-ol)= n*MM = (27/700)*88.146 = 3.40 (2s.f.)

 

[bedpotato]

If one mole of pentanol is combusted, 2800kJ of heat is generated. How many moles of pentanol should be combusted to generate 108kJ of heat?
1mol ---- 2800kJ
xmol ---- 108kJ

x = 108kJ . 1mol / 2800kJ
= 0.03857...mol

Then just use m = n x M, and you should get D.

 

[Jimmy2064]

Way over thought it, thanks heaps guys!