Calculate the change in enthalpy associated with the combustion of 322 g of ethanol. C2H5OH(l)+3O2(g)⟶2CO2(g)+3H2O(l)ΔH∘ c=−1366.8kJ/mol

The change in enthalpy associated with the combustion of 322 g of ethanol is

Explanation:

To calculate the number of moles we use the equation:

The balanced chemical reaction is:

Given :

Energy released when 1 mole of ethanol is combusted = 1366.8 J

Thus Energy released when 7 moles of ethanol is combusted =

Thus the change in enthalpy associated with the combustion of 322 g of ethanol is

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml

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the scientist could experiment with moth repellent or oter ways to keep the moths away, but that would not answer his original question.