Question

When you see the tip of a match fire, the chemical reaction is likely to be
P4S3+802 gives P4O10 +3SO2
What is the minimum amount of P4S3 that would have to be burned to produce at least 1.0 g of P4O10 and at
least 1.0 g of SO2​

Answer 1

Given:

Molar mass of  $P_{4}$$O_{10}$ = 283.886 g/mol

Molar mass of $SO_{2}$ = 64 g/mol

Molar mass of $P_{4} S_{3}$ = 220.093 g/mol

To find:

The minimum amount of  $P_{4} S_{3}$ required to produce at least 1 gram of each product

Soluion:

The given balanced chemical reaction is;

          $P_{4} S_{3}$ + 8$O_{2}$ → $P_{4} O_{10}$ + 3$SO_{2}$

• By stoichiometry of the reaction we conclude that;

        If, 283.886 grams of $P_{4} O_{10}$ is produced by 220.093 grams of $P_{4} S_{3}$

        Then, 1 gram of $P_{4} O_{10}$ will be produced by $\frac{220.093}{283.886}$ × 1g = 0.786g of $P_{4} S_{3}$

• By stoichiometry of the reaction we conclude that;

         If, 64 grams of $SO_{2}$ is produced by 220.093 grams of $P_{4}S_{3}$

         Then, 1 gram of $SO_{2}$ will be produced by $\frac{220.093}{64}$ × 1g = 3.439g of $P_{4} S_{3}$

Therefore, the minimum amount of  $P_{4} S_{3}$  required to produce at least 1 gram of each product is 0.786 grams.