Respuesta :
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.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ....(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:
[tex]\text{Moles of potassium chlorate}=\frac{56.0g}{122.55g/mol}=0.456mol[/tex]
For the given chemical reaction:
[tex]10KClO_3+12P\rightarrow 3P_4O_{10}+10KCl[/tex]
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 = [tex]\frac{3}{10}\times 0.456=0.136moles[/tex] 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:
[tex]0.136mol=\frac{\text{Mass of tetraphosphorus decoxide}}{283.886g/mol}\\\\\text{Mass of tetraphosphorus decoxide}=38.60g[/tex]
- 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 = [tex]\frac{10}{10}\times 0.456=0.456moles[/tex] 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:
[tex]0.456mol=\frac{\text{Mass of potassium chloride}}{74.55g/mol}\\\\\text{Mass of potassium chloride}=33.99g[/tex]
- 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 = [tex]\frac{12}{10}\times 0.456=2.631moles[/tex] 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:
[tex]2.631mol=\frac{\text{Mass of red phosphorus}}{30.97g/mol}\\\\\text{Mass of red phosphorus}=81.48g[/tex]
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.
