Question

The reaction between potassium chlorate and red phosphorus takes place when you strike amatch on a matchbox. If you were to react$\mathbf{52}\mathbf{.}\mathbf{9}\mathbf{\text{g}}$of potassium chlorate$\left({\text{KClO}}_3\right)$with excess redphosphorus, what mass of tetra phosphorus decoxiderole="math" localid="1648562536382" $\left({\text{P}}_4{\text{O}}_{10}\right)$could be produced?

role="math" localid="1648562558748" ${\mathbf{\text{KClO}}}_{\mathbf{3}}\left(\text{s}\right){\mathbf{\text{+P}}}_{\mathbf{4}}\left(\text{s}\right){\mathbf{\text{→P}}}_{\mathbf{4}}{\mathbf{\text{O}}}_{\mathbf{10}}\left(\text{s}\right)\mathbf{\text{+KCl}}\left(\text{s}\right)$ (Unbalanced).

Short answer

The mass of tetra phosphorus decoxide$\left({\text{P}}_4{\text{O}}_{10}\right)$ could be produced is$36.9\text{g}$ .

Step-by-step solution

Definition

The mass of any material expressed in atomic mass units isquantitatively equivalent to its molar mass in grams per mole.

Balanced Equation 

The given equation is balanced as:

${\text{10KClO}}_3\left(\text{s}\right){\text{+3P}}_4\left(\text{s}\right){\text{→3P}}_4{\text{O}}_{10}\left(\text{s}\right)\text{+10KCl}$

Determining the moles of potassium chlorate

From the given,

Mass of potassium chlorate $=52.9\text{g}$

Molar mass ofrole="math" localid="1648560800416" ${\mathbf{\text{KClO}}}_{\mathbf{3}}\mathbf{\text{=MassofCl+MassofK+3}}\left(\text{MassofC}\right)$

role="math" localid="1648562033167" $$\begin{gathered} =32.10\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.+35.45\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.+3\left(16.00\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\right) \\ =122.55\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right. \end{gathered}$$

$\mathbf{\text{Molesofpotassiumchlorate=}}\frac{\mathbf{\text{Mass}}}{\mathbf{\text{Molarmass}}}$

$=\frac{52.9\text{g}}{122.55{\displaystyle \raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.}}$

$=0.432\text{g}$

Therefore,the moles ofpotassium chlorate are$\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{432}\mathbf{}\mathbf{}\mathbf{\text{g}}$.

Determining the mass of tetra phosphorus decoxide produced

From the given equation,10 moles of potassium chlorate give 10 moles of tetra phosphorus decoxide.

So,

$=\frac{0.432\times 3}{10}$
$=0.13$Moles of tetra phosphorus decoxide

Molar mass of tetra phosphorus decoxide =$\mathbf{4}\mathbf{\times }\left(\text{MassofP}\right)\mathbf{+}\mathbf{}\mathbf{10}\mathbf{}\left(\text{MassofO}\right)$

$$\begin{gathered} =4\left(32.10\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\right)+10\left(16.00\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\right) \\ =283.88\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right. \end{gathered}$$

Mass of tetra phosphorus decoxide produced$\mathbf{=}\mathbf{\left(}\mathbf{\text{Mass}}\mathbf{\right)}\mathbf{\times }\mathbf{\left(}\mathbf{\text{Molarmass}}\mathbf{\right)}$

$=0.13\times 283.88\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$

Therefore, the mass of tetra phosphorus decoxideproduced is $\mathbf{36}\mathbf{.}\mathbf{9}\mathbf{\text{g}}$.