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hi year 11 chem help pls, exam tomorrow. (1 Viewer)

jazz519

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Two formulas are used in these questions:
q = mcΔT

ΔHc = -q/n(fuel)

Always first should assign the values in the question to variables:
m(C2H5OH) = 1.275 g ==> n(C2H5OH) = m/MM = 1.275 / 46.068 = 0.02767.. moles

m = 100 g = 0.1 kg
ΔT= 14.4 C
c = 4.18 J/g/C
ΔHc = -1364 kJ/mol (this should be negative in the data given because it's exothermic, so question has a slight formatting error)

Start by choosing the formula which has 1 unknown variable and use this to do the calculation:
this case it is the q = mcΔT as only unknown is the q

q = (0.1)(4.18)(14.4) = 6.0192 kJ

After use the other formula:
ΔHc = -q/n(fuel)
ΔHc = -6.0192/0.02767...
ΔHc = 217.535.. kJ/mol

Normally it will end here, but what makes this question different is they want heat absorbed by the water:

%heat absorbed = (ΔHc experiment / ΔHc theoretical) x 100
%heat absorbed = (217.535... / 1364) x 100
%heat absorbed = 15.948...
%heat absorbed = 15.9% (3 sig figs)
 

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