Question

P4s3 + 8o2 p4o10 + 3so2 calculate minimum mass of p4s3 is required to produce atleast 1 gm of each product

Answer 1

Answer: The minimum amount of $P_4S_3$ required to produced at least 1 gram of each product is 0.786 grams.

Explanation : Given,

Molar mass of $P_4O_{10}$ = 283.886 g/mol

Molar mass of $SO_2$ = 64 g/mol

Molar mass of $P_4S_3$ = 220.093 g/mol

The given balanced chemical reaction is,

$P_4S_3+8O_2\rightarrow P_4O_{10}+3SO_2$

By stoichiometry of the reaction we conclude that,

If, 283.886 grams of $P_4O_{10}$ is produced by 220.093 grams of $P_4S_3$

Then, 1 gram of $P_4O_{10}$ will be produced by $\frac{220.093}{283.886}\times 1g=0.786g$ of $P_4S_3$

By stoichiometry of the reaction we conclude that,

If, 64 grams of $SO_{2}$ is produced by 220.093 grams of $P_4S_3$

Then, 1 gram of $SO_{2}$ will be produced by $\frac{220.093}{64}\times 1g=3.439g$ of $P_4S_3$

Therefore, the minimum amount of $P_4S_3$ required to produced at least 1 gram of each product is 0.786 grams.

Answer 2

Explanation:

Molar mass of P_4O_{10}P

4

O

10

= 283.886 g/mol

Molar mass of SO_2SO

2

= 64 g/mol

Molar mass of P_4S_3P

4

S

3

= 220.093 g/mol

The given balanced chemical reaction is,

P_4S_3+8O_2\rightarrow P_4O_{10}+3SO_2P

4

S

3

+8O

2

→P

4

O

10

+3SO

2

By stoichiometry of the reaction we conclude that,

If, 283.886 grams of P_4O_{10}P

4

O

10

is produced by 220.093 grams of P_4S_3P

4

S

3

Then, 1 gram of P_4O_{10}P

4

O

10

will be produced by \frac{220.093}{283.886}\times 1g=0.786g

283.886

220.093

×1g=0.786g of P_4S_3P

4

S

3