Question

What weights of ${\mathrm{P}}_4{\mathrm{O}}_6$ and ${\mathrm{P}}_4{\mathrm{O}}_{10}$ will be produced by the combustion of $31\mathrm{g}$ of ${\mathrm{P}}_4$ in $32\mathrm{g}$ of oxygen leaving no ${\mathrm{P}}_4$ and no ${\mathrm{O}}_2$?

• A. $55\mathrm{g}$ and $71\mathrm{g}$
• B. $17.5\mathrm{g}$ and $19.5\mathrm{g}$
• C. $2.75\mathrm{g}$ and $35.5\mathrm{g}$
• D. $2.75\mathrm{g}$ and $29.5\mathrm{g}$

Answer 1

The correct option is C

$2.75\mathrm{g}$ and $35.5\mathrm{g}$

Explanation for the correct option:

Step 1: Write the balanced chemical reaction and molar mass of ${\mathrm{P}}_4$ and ${\mathrm{O}}_2$.

${\mathrm{P}}_4\left(s\right)+3{\mathrm{O}}_2\left(g\right)\to {\mathrm{P}}_4{\mathrm{O}}_6\left(s\right)$

Molar mass of ${\mathrm{P}}_4=124\mathrm{g}/\mathrm{mol}$

Molar mass of ${\mathrm{O}}_2=32\mathrm{g}/\mathrm{mol}$

Step 2: Calculate the moles of ${\mathrm{P}}_4$ and ${\mathrm{O}}_2$.

Moles of ${\mathrm{P}}_4=\frac{31\mathrm{g}}{124\mathrm{g}/\mathrm{mol}}=0.25\mathrm{moles}$

Moles of ${\mathrm{O}}_2=\frac{32\mathrm{g}}{32\mathrm{g}/\mathrm{mol}}=1\mathrm{mole}$

Step 3: Determine the limiting reagent in the reaction.

As $1\mathrm{mole}$ of ${\mathrm{P}}_4$ reacts with $3\mathrm{moles}$ of ${\mathrm{O}}_2$

Therefore,

$0.25\mathrm{moles}$ of ${\mathrm{P}}_4$ will react with $0.25\times 3\mathrm{moles}=0.75\mathrm{moles}$ of ${\mathrm{O}}_2$

Thus, ${\mathrm{P}}_4$ is the limiting reagent which is completely consumed in the reaction and Oxygen is left.

Step 4: Determine the moles of ${\mathrm{P}}_4{\mathrm{O}}_6$

Since $1\mathrm{mole}$ of ${\mathrm{P}}_4$ produces $1\mathrm{mole}$ of ${\mathrm{P}}_4{\mathrm{O}}_6$, therefore, $0.25\mathrm{moles}$ of ${\mathrm{P}}_4$ produces $0.25\mathrm{moles}$ of ${\mathrm{P}}_4{\mathrm{O}}_6$.

Step 5: Write the reaction of combustion of ${\mathrm{P}}_4{\mathrm{O}}_6$ and find the limiting reagent of the reaction.

${\mathrm{P}}_4{\mathrm{O}}_6\left(s\right)+2{\mathrm{O}}_2\left(g\right)\to {\mathrm{P}}_4{\mathrm{O}}_{10}\left(s\right)$

As $1\mathrm{mole}$ of ${\mathrm{P}}_4{\mathrm{O}}_6$ reacts with $2\mathrm{moles}$ of ${\mathrm{O}}_2$

Therefore,

$0.25\mathrm{moles}$ of ${\mathrm{P}}_4{\mathrm{O}}_6$ will react with $0.25\times 2\mathrm{moles}=0.50\mathrm{moles}$ of ${\mathrm{O}}_2$

The amount of Oxygen required for this reaction is not available as $0.75\mathrm{moles}{\mathrm{O}}_2$ is already consumed and only $0.25\mathrm{moles}{\mathrm{O}}_2$ is left, Oxygen is the limiting reagent.

Step 6: Calculate the moles of ${\mathrm{P}}_4{\mathrm{O}}_{10}$ produced.

Since $2\mathrm{moles}$ of ${\mathrm{O}}_2$ produces $1\mathrm{mole}$ of ${\mathrm{P}}_4{\mathrm{O}}_{10}$, therefore, $0.25\mathrm{moles}$ of ${\mathrm{O}}_2$ produces $\frac{0.25}{2}=0.125\mathrm{moles}$ of ${\mathrm{P}}_4{\mathrm{O}}_{10}$.

Number of moles of ${\mathrm{P}}_4{\mathrm{O}}_{10}$ left $=0.25-0.125=0.125\mathrm{moles}$

Step 7: Calculate the weight of ${\mathrm{P}}_4{\mathrm{O}}_6$ and ${\mathrm{P}}_4{\mathrm{O}}_{10}$

Mass of ${\mathrm{P}}_4{\mathrm{O}}_6=0.125\mathrm{mol}\times 220\mathrm{g}/\mathrm{mol}=27.5\mathrm{g}$

Mass of ${\mathrm{P}}_4{\mathrm{O}}_6=0.125\mathrm{mol}\times 220\mathrm{g}/\mathrm{mol}=27.5\mathrm{g}$

Hence, the correct option is (C).

Explanation for the incorrect options:

Since, the correct weights of ${\mathrm{P}}_4{\mathrm{O}}_6$ and ${\mathrm{P}}_4{\mathrm{O}}_{10}$ are $2.75\mathrm{g}$ and $35.5\mathrm{g}$. Therefore, options (A), (B) and (D) stand incorrect.

Therefore, the correction option is (C).