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Iron(III) oxide reacts with carbon monoxide according to the equation: Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g) A reaction mixture initially contains 22.00 g Fe2O3 and 14.10 g CO. Once the reaction has occurred as completely as possible, what mass (in g) of the excess reactant is left?

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lamnd0408

Answer:

mass of excessive CO = 2.55 gram

Explanation:

Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g)

moles of Fe2O3 = mass / formula mass = 22.00/(56x2+16x3)=0.1375 (mol)

moles of CO = mass / formula mass = 14.1/(16+12) = 0.503

Fe2O3 reacts completely meanwhile CO is excessive.

mass of CO reacts = 3 x nFe2O3 x M = 3 x 0.1375 x (16+12) = 11.55 gram

mass of excessive CO = initial mass - reacted mass = 14.1 - 11.55 = 2.55 gram

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