Question

The following thermochemical equation is for the reaction of iron(III) oxide(s) with hydrogen(g) to form Fe304(s) and water(g
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Answer #1

Given that

3Fe2O3(s) +H2(g) ----- > 2 Fe3O4 (s) + H2O (g)

dH = -6.00 KJ

given that energy which is produce = 0.778 kj

Number of moles = 0.778 KJ * 3 mole Fe2O3 / 6.00 KJ

=0.389 mole Fe2O3

Amount in g = number of moles * molar mass

=0.389 mole Fe2O3*159.69 g/mol

= 62.1 g Fe2O3

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