Question #2684

1.6 g ( an excess) of Mg was added to 100 cm[sup]3[/sup] of 2.0 mol dm[sup]–3[/sup] CuSO4(aq). The temperature rose from 20C to 65C (45’C rise). Find ΔHr for the reaction: (c = 4.2 J g[sup]-1[/sup] K[sup]-1[/sup])

Mg(s) + CuSO[sub]4[/sub](aq) --> MgSO[sub]4[/sub](aq) + Cu(s)

Find the energy change by using ΔH = - mcΔT

(MY ANSWER) 100 x 4.2 x 45 = 18900 J mol[sup]-1[/sup] = 18.9 kJ mol[sup]-1[/sup] (which is correct)

Find how many moles reacted (n)

(MY ANSWER) 1.6/159.6 = 0.01 then 0.01/2 = 0.005 (which is incorrect, correctly should be 2x100/1000; do not know how!)

Scale the quantities for the combustion of 1 mole (ΔH/n)

18.9/0.005 = -3780 kJ mol[sup]-1[/sup] (using the previous incorrect figure this is also wrong)

Could you please help to explain step 2?

Mg(s) + CuSO[sub]4[/sub](aq) --> MgSO[sub]4[/sub](aq) + Cu(s)

Find the energy change by using ΔH = - mcΔT

(MY ANSWER) 100 x 4.2 x 45 = 18900 J mol[sup]-1[/sup] = 18.9 kJ mol[sup]-1[/sup] (which is correct)

Find how many moles reacted (n)

(MY ANSWER) 1.6/159.6 = 0.01 then 0.01/2 = 0.005 (which is incorrect, correctly should be 2x100/1000; do not know how!)

Scale the quantities for the combustion of 1 mole (ΔH/n)

18.9/0.005 = -3780 kJ mol[sup]-1[/sup] (using the previous incorrect figure this is also wrong)

Could you please help to explain step 2?

Expert's answer

Mass of magnesium given to excess. Therefore, to calculate the amount of matter we have to use the amount of copper sulfate.

n(CuSO_{4}) = (2 mol/dm^{3})*100cm^{3}/1000 = 0.005 mol

n(CuSO

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