Question

1. iron is extracted from the mineral magnetite, fe₃o₄:

fe₃o₄(s) + 2 c(s) → 3 fe(s) + 2 co₂(g)
a 35.0 g sample of magnetite produces 15.0 g of iron.
determine the percentage yield of this reaction. t/i

Explanation:

Step1: Calculate the molar mass of \(Fe_3O_4\)

The molar mass of \(Fe\) is \(55.85\space g/mol\), and the molar mass of \(O\) is \(16.00\space g/mol\).

$$M(Fe_3O_4)=3\times55.85 + 4\times16.00=231.55\space g/mol$$

Step2: Calculate the moles of \(Fe_3O_4\)

Using the formula \(n=\frac{m}{M}\), where \(m = 35.0\space g\) and \(M = 231.55\space g/mol\)

$$n(Fe_3O_4)=\frac{35.0}{231.55}\approx0.151\space mol$$

Step3: Calculate the theoretical moles of \(Fe\)

From the balanced equation \(Fe_3O_4(s)+2C(s)\to3Fe(s)+2CO_2(g)\), the mole ratio of \(Fe_3O_4\) to \(Fe\) is \(1:3\).
So, \(n_{theo}(Fe)=3\times n(Fe_3O_4)=3\times0.151 = 0.453\space mol\)

Step4: Calculate the theoretical mass of \(Fe\)

Using \(m = n\times M\), with \(n = 0.453\space mol\) and \(M(Fe)=55.85\space g/mol\)

$$m_{theo}(Fe)=0.453\times55.85\approx25.3\space g$$

Step5: Calculate the percentage yield

The formula for percentage yield is \(\%\text{yield}=\frac{m_{actual}}{m_{theo}}\times100\%\)
Given \(m_{actual}=15.0\space g\) and \(m_{theo}\approx25.3\space g\)

$$\%\text{yield}=\frac{15.0}{25.3}\times100\%\approx59.3\%$$

Answer:

The percentage yield of the reaction is approximately \(59.3\%\)