0.800g Match to SO?: Volume Calculation at 725 Torr & 32°C!

Step 1 of 2

Here, we are going to calculate the volume of sulfur dioxide which is produced in the combustion of tetraphosphorus trisulfide.

Calculation of the moles of the reactant and product.

The balanced chemical equation is given below:

\(\mathrm{P}_{4} S_{3}(s)+8 \mathrm{O}_{2}(g) \rightarrow \mathrm{P}_{4} \mathrm{O}_{10}(g)+3 \mathrm{SO}_{2}(g)\)

The mass of \(\mathrm{P}_{4} \mathrm{~S}_{3}=0.800 \mathrm{~g}\)

The molar mass of \(\mathrm{P}_{4} \mathrm{~S}_{3}=220.09 \mathrm{~g} / \mathrm{mol}\)

Therefore, the moles of \(\mathrm{P}_{4} \mathrm{~S}_{3}=0.800 \mathrm{~g} \mathrm{P}_{4} \mathrm{~S}_{3} \times \frac{1.0 \mathrm{~mol} \mathrm{P}_{4} S_{3}(s)}{220.09 \mathrm{~g} \mathrm{P}_{4} S_{3}(s)}\)

                                                        \(=3.63 \times 10^{-3} \mathrm{~mol}\)

Thus, the moles of \(\mathrm{SO}_{2}(\mathrm{~g})\)

\(3.63 \times 10^{-3} \mathrm{molP}_{4} \mathrm{~S}_{3}(s) \times \frac{3.0 \mathrm{molsO}_{2}(g)}{1.0 \mathrm{molP}_{4} \mathrm{~S}_{3}(s)}=0.0109 \mathrm{~mol} \mathrm{SO}_{2}(g)\)