Question

What is the weight Of ${\mathrm{V}}_2{\mathrm{O}}_5$ produced from $2g$of $\mathrm{VO}$ and $5.75g$ of ${\mathrm{Fe}}_2{\mathrm{O}}_3$?

• A. $5.07g$
• B. $5.17g$
• C. $5.27g$
• D. $5.37g$

Answer 1

The correct option is B

$5.17g$

The explanation of the correct option:

B)$\mathbf{5}\mathbf{.}\mathbf{17}\mathbf{}\mathit{g}$

Step 1: Write the balanced chemical equation.

• $2\mathrm{VO}\left(\mathrm{s}\right)+{\mathrm{Fe}}_2{\mathrm{O}}_3\left(\mathrm{s}\right)\to 6\mathrm{FeO}\left(\mathrm{s}\right)+{\mathrm{V}}_2{\mathrm{O}}_5\left(\mathrm{s}\right)$

Step 2: Calculate the molecular formula

• Atomic masses of $\mathrm{V}=50.9\mathrm{u};\mathrm{Fe}=55.8\mathrm{u};\mathrm{O}=16\mathrm{u}$
• $$\begin{gathered} 2\mathrm{VO}=2(50.9+16) \\ =133.8\mathrm{g} \end{gathered}$$

• $$\begin{gathered} 3{\mathrm{Fe}}_2{\mathrm{O}}_3=3(55.8\times 2+16\times 3) \\ =479\mathrm{g} \end{gathered}$$

• $$\begin{gathered} 6\mathrm{FeO}=6(55.9+16) \\ =431.4\mathrm{g} \end{gathered}$$

Step 3: Calculation

• Therefore, $133.8\mathrm{g}\mathrm{of}\mathrm{VO}\mathrm{is}\mathrm{required}\mathrm{for}479\mathrm{g}\mathrm{of}{\mathrm{Fe}}_2{\mathrm{O}}_3$ so, $2\mathrm{g}\mathrm{of}\mathrm{VO}\mathrm{requires}$

$\frac{479\times 2}{133.8}=7.15g$

• But it is given that $5.75\mathrm{g}\mathrm{of}{\mathrm{Fe}}_2{\mathrm{O}}_3$and hence it is a limiting reagent.
• So from $479\mathrm{g}\mathrm{of}{\mathrm{Fe}}_2{\mathrm{O}}_3$, $431.4\mathrm{g}\mathrm{of}\mathrm{FeO}$is obtained. Therefore, from $5.75\mathrm{g}\mathrm{of}\mathrm{FeO}$ we get

$\frac{431.4\times 5.75}{479}=5.17g$

Hence, option C) $\mathbf{5}\mathbf{.}\mathbf{17}\mathbf{}\mathit{g}$ is correct.