Question

Calculate the work done when 50.1043 g of iron reacts with hydrochloric acid to produce FeCl2 and hydrogen in an open beaker on the lab bench at 23.41 degrees C. Assume hydrogen gas behaves ideally. Fe(s) + 2 HCl(Aqua)--> FeCl2(aq) + H2(g)

Not sure how to approach this problem so all steps would be helpful!!

Answer 1

When a gas is formed during the reaction then we give work done in form of ideal gas equation, i.e

work done=  -ΔnRT

R=gas constant=8.314

delta n= number of moles of H2 formed

T=temp. = 23.41+271= 296.4 K

Now, determining the no. of moles of H2 formed:

moles of Fe present= 50.1043/55.845 = 0.897 moles

according to the stoichiometric coefficients, one mole of Fe forms one moles of H2.

SO 0.897 moles of Fe forms 0.897 moles of H2.

therefore,

work done= -0.897*296.4*8.314 = -2210.45 J