Question

The reaction between potassium chlorate and red phosphorus takes place when you strike a match on a matchbox. If you were to react 93.3 g of potassium chlorate with excess red phosphorus (P4), to produce tetraphosphorus decaoxide and potassium chloride, what mass of tetraphosphorus decaoxide would be produed?

Answer 1

Answer: The mass of tetraphosphorus decaoxide formed is 64.81g

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$   ....(1)

• For potassium chlorate:

Given mass of potassium chlorate = 93.3 g

Molar mass of potassium chlorate = 122.55 g/mol

Putting values in above equation, we get:

$\text{Moles of potassium chlorate}=\frac{93.3g}{122.55g/mol}=0.761mol$

For the given chemical reaction:

$10KClO_3+12P\rightarrow 3P_4O_{10}+10KCl$

Red phosphorus is given in excess . So, it is considered as an excess reagent and potassium chlorate is considered as a limiting reagent.

By Stoichiometry of the reaction:

10 moles of potassium chlorate reacts with 3 moles of tetraphosphorus decaoxide

So, 0.761 moles of potassium chlorate will react with = $\frac{3}{10}\times 0.761=0.2283moles$ of tetraphosphorus decaoxide

Calculating the mass of tetraphosphorus decaoxide by using equation 1, we get:

Molar mass of tetraphosphorus decaoxide = 283.886 g/mol

Moles of tetraphosphorus decaoxide = 0.2283 moles

Putting values in equation 1, we get:

$0.2283mol=\frac{\text{Mass of tetraphosphorus decaoxide}}{283.886g/mol}\\\\\text{Mass of tetraphosphorus decaoxide}=64.81g$

Hence, the mass of tetraphosphorus decaoxide formed is 64.81g