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Please its urgent! 8P4 + 3S8 -> 8P4S3

The molar masses of the substances are as follows: P4 = 123.895g/mol, S8 = 256g/mol, P4S3 = 220.093g/mol

If 20.0g of P4 and 40.0g of S8 are combined, what is the amount of P4S3 produced in moles?

User Mmgp
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1 Answer

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Answer:

The amount of P₄S₃ produced is 0.16 moles.

Step-by-step explanation:

The balanced reaction is:

8 P₄ + 3 S₈ → 8 P₄S₃

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate:

  • P₄: 8 moles
  • S₈: 3 moles
  • P₄S₃: 8 moles

Being the molar mass of each compound:

  • P₄: 123.895 g/mol
  • S₈: 256 g/mol
  • P₄S₃: 220.093 g/mol

then by stoichiometry of the reaction, the following amounts of reactant and product participate in the reaction:

  • P₄: 8 moles* 123.895 g/mol= 991.16 g
  • S₈: 3 moles* 256 g/mol= 768 g
  • P₄S₃: 8 moles* 220.093 g/mol= 1,760.744 g

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

In this case, you calculate the limiting reactant using the following rule of three: if by stoichiometry 768 g of S₈ react with 991.16 g of P₄, 40 g of S₈ with how much mass of P₄ will it react?


mass of P_(4)=(40 grams of S_(8) *991.16 grams of P_(4) )/(768 grams of S_(8))

mass of P₄= 51.62 grams

But 51.62 grams of P₄ are not available, 20 grams are available. Since you have less mass than you need to react with 40 grams of S₈, P₄ will be the limiting reagent.

Then you can apply the following rule of three: if by stoichiometry 991.16 grams of P₄ form 8 moles of P₄S₃, 20 grams of P₄ will form how many moles of P₄S₃?


moles of P_(4) S_(3)=(20 grams of P_(4) *8 moles ofP_(4) S_(3) )/(991.16 grams of P_(4))

moles of P₄S₃= 0.16

The amount of P₄S₃ produced is 0.16 moles.

User Rohan Sampat
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