Question

The reaction between potassium chlorate (KCIO,) and red phosphorus (P.) takes place when one strikes a match. The products of the reaction are tetraphosphorus decoxide and potassium chloride. If 56.0 grams of KCIO, are reacted with an excess amount of red phosphorus, how many grams of P0o and KCI can be produced? How much red phosphorus is consumed in the reaction? (15 pts) Write the balanced reaction first!

Answer 1

Answer: Amount of tetraphosphorus decoxide formed is 38.60g, amount of potassium chloride formed is 33.99g and amount of red phosphorus consumed in the reaction is 81.48 g.

Step-by-step 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 = 56.0 g

Molar mass of potassium chlorate = 122.55 g/mol

Putting values in above equation, we get:

`\text{Moles of potassium chlorate}=(56.0g)/(122.55g/mol)=0.456mol`

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.

• For tetraphosphorus decoxide:

By Stoichiometry of the reaction:

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

So, 0.456 moles of potassium chlorate will react with =
`(3)/(10)* 0.456=0.136moles` of tetraphosphorus decoxide

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

Molar mass of tetraphosphorus decoxide = 283.886 g/mol

Moles of tetraphosphorus decoxide = 0.136 moles

Putting values in equation 1, we get:

`0.136mol=\frac{\text{Mass of tetraphosphorus decoxide}}{283.886g/mol}\\\\\text{Mass of tetraphosphorus decoxide}=38.60g`

• For potassium chloride:

By Stoichiometry of the reaction:

10 moles of potassium chlorate reacts with 10 moles of potassium chloride

So, 0.456 moles of potassium chlorate will react with =
`(10)/(10)* 0.456=0.456moles` of potassium chloride

Calculating the mass of potassium chloride by using equation 1, we get:

Molar mass of potassium chloride = 74.55 g/mol

Moles of potassium chloride = 0.456 moles

Putting values in equation 1, we get:

`0.456mol=\frac{\text{Mass of potassium chloride}}{74.55g/mol}\\\\\text{Mass of potassium chloride}=33.99g`

• For Red phosphorus (excess reagent)

By Stoichiometry of the reaction:

10 moles of potassium chlorate reacts with 12 moles of red phosphorus.

So, 0.456 moles of potassium chlorate will react with =
`(12)/(10)* 0.456=2.631moles` of red phosphorus

Calculating the mass of red phosphorus by using equation 1, we get:

Molar mass of red phosphorus = 30.97 g/mol

Moles of red phosphorus = 2.631 moles

Putting values in equation 1, we get:

`2.631mol=\frac{\text{Mass of red phosphorus}}{30.97g/mol}\\\\\text{Mass of red phosphorus}=81.48g`

Hence, amount of tetraphosphorus decoxide formed is 38.60g, amount of potassium chloride formed is 33.99g and amount of red phosphorus consumed in the reaction is 81.48 g.