Question

40g of molten iron (II) oxide react with 10g of magnesium according to the following equation:
FeO(l)+Mg(l)--> Fe(l)+MgO(s)
(A) what is the mass of iron produced?
(B) if the actual yield of Fe is 18.8g,calculate the percentage yield of this reaction.

Answer 1

Answer on Question #56618 - Chemistry - Inorganic Chemistry

Question:

40g of molten iron (II) oxide react with 10g of magnesium according to the following equation:

$$\mathrm{FeO(I)} + \mathrm{Mg(I)} \rightarrow \mathrm{Fe(I)} + \mathrm{MgO(s)}$$

(A) what is the mass of iron produced?

(B) if the actual yield of Fe is 18.8g, calculate the percentage yield of this reaction.

Solution:

$$\mathrm{MW_{FeO}} = 71.844\ \mathrm{g\ mol^{-1}};$$ $$\mathrm{MW_{Mg}} = 24.305\ \mathrm{g\ mol^{-1}};$$ $$\mathrm{MW_{Fe}} = 55.845\ \mathrm{g\ mol^{-1}};$$ $$n = \mathrm{m/MW};$$ $$n_{\mathrm{FeO}} = 40\ \mathrm{g}/71.844\ \mathrm{g\ mol^{-1}} = 0.5568\ \mathrm{mol};$$ $$n_{\mathrm{Mg}} = 10\ \mathrm{g}/24.305\ \mathrm{g\ mol^{-1}} = 0.4114\ \mathrm{mol};$$

Mg is a limiting reagent:

$$n_{\mathrm{Fe}} = n_{\mathrm{Mg}} = 0.4114\ \mathrm{mol};$$ $$m_{\mathrm{Fe}} = n_{\mathrm{Fe}} \cdot \mathrm{MW_{Fe}} = 0.4114\ \mathrm{mol} \cdot 55.845\ \mathrm{g\ mol^{-1}} = 22.98\ \mathrm{g} \approx 23\ \mathrm{g}$$

Yield of Fe:

$$W_{\mathrm{Fe}} = 18.8\ \mathrm{g}/22.98\ \mathrm{g} = 81.82\% \approx 82\%$$

Answer:

(A) 23 g

(B) 82%

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