Question

What mass of ${P_2}{O_6}$ will be produced by the combustion of $2.0\,g$of ${P_4}$ with $2.0\,g$ of ${O_2}$ ?
A. $0.0145\,\,mol$
B. $0.072\,\,mol$
C. $0.029\,\,mol$
D. $0.0048\,\,mol$

Answer 1

In order to answer this question, we should know about the mole concept. Here, the mass of ${P_4}$ and ${O_2}$ is given which helps in determining the number of moles of phosphorus and oxygen. Also, we should know about the chemical reaction taking place during the combustion process.

Complete answer:
Let us understand the answer to this question in complete detail.
Here we have given the weight of phosphorus and oxygen. So, we calculate the number of moles each.
We know that,
$Number\,\,of\,moles = \dfrac{{Given\,\,weight(g)}}{{Molar\,\,mass\,(g/mol)}}$
Number of moles of ${P_4}$ = $\dfrac{{2.0}}{{31 \times 4}}$ = $0.016$ mol
Number of moles of ${O_2}$= $\dfrac{{2.0}}{{16 \times 2}}$= $0.1625$ mol
The chemical reaction that taking place during combustion is as follows:
${P_4}\, + 3{O_2} \to {P_4}{O_6}$
Here, by looking at the chemical reaction we observed that one mole of phosphorus (${P_4}$) reacts with three moles of oxygen (${O_2}$) to produce one mole of ${P_4}{O_6}$.
So, if $0.016$ mol of ${P_4}$ reacts with $3 \times 0.1625\,mol$ of oxygen then it produces $0.016$ mol of ${P_4}{O_6}$.
Moles of oxygen = $3 \times 0.1625\,mol$= $0.048$ mol is used to ${O_2}$ to make ${P_4}{O_6}$
Moles of ${O_2}$ unreacted = $0.0625 - 0.048$= $0.0145$ mol of ${O_2}$
This ${O_2}$ can react according to the reaction given below:
${P_4}{O_6} + 2{O_2} \to {P_4}{O_{10}}$
Now,
$(0.0415\,\,mol\,{O_2})\left( {\dfrac{{1\,mol\,{P_4}{O_{10}}}}{{2\,mol\,{O_2}}}} \right) = 0.00721\,mol\,{P_4}{O_{10}}$
The originally formed ${P_4}{O_6}$ was $0.016$ mol in which $0.00721$is converted to ${P_4}{O_{10}}$.
So, ($0.016 - 0.00721$)=$0.009\,$mol of ${P_4}{O_6}$

The correct answer is option (B).

Additional information:
Now, we can calculate the mass of ${P_4}{O_6}$ produced.
Let’s first calculate the molar mass of ${P_4}{O_6}$
${P_4}{O_6}$= $(4 \times 31) + (16 \times 6)$
${P_4}{O_6}$= $122 + 96$=$220$g/mol
Now, again we use this formula:
$Number\,\,of\,moles = \dfrac{{Given\,\,weight(g)}}{{Molar\,\,mass\,(g/mol)}}$
$Given\,\,weight(g)\, = \,Molar\,\,mass\,(g/mol)\, \times \,Number\,\,of\,\,moles$
Given weight (g) = $220 \times 0.009\,$= 1.98\,g$$$${P_4}{O_6}

Note:
We must remember the formula used in the above question as it is very useful in the calculation for microscopic particles. Also, we should know the atomic mass as well as the molecular mass of the elements given.