Question

p4s3 + 02= p4o10+so2 the mass of p4o10 produced if 440 gm of p4s3 is mixed with 384 gm of o2.

Answer 1

Answer is 664 grams. Since, O2 has molar mass as 32g therefore, 384 gram contains 12 moles of O2. Thus, S has molar mass as 32 since it is 3 moles in P4S3 therefore molar mass of P4 would be added to o10 that is 320. That gives 344+ 320= 664

Answer 2

Answer: 426 g

Explanation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of $P_4S_3$

Given mass of  $P_4S_3$= 440 grams

Molar mass of  $P_4S_3$= g/mol

$\text{Number of moles}=\frac{440g}{220g/mol}=2moles$

Moles of $O_2$

Given mass of $O_2$ = 384 g

Molar mass of $O_2$ = 32 g/mol

$\text{Number of moles}=\frac{384g}{32g/mol}=12moles$

For the given chemical reaction, the balanced equation is

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

By Stoichiometry:

8 moles of $O_2$ react with 1 mole of  $P_4S_3$

So, 12 moles of $O_2$ react with = $\frac{1}{8}\times 12=1.5moles$  $P_4S_3$

$O_2$ is considered as a limiting reagent because it limits the formation of product and $P_4S_3$ is the excess reagent.

8 moles 8 moles of $O_2$ produce = 1 mole of $P_4O_{10}$

So, 12 moles of $O_2$ will produce= $\frac{1}{8}\times 12=1.5moles$ of $P_4O_{10}$

Mass of $P_4O_{10}=moles\times Molar mass=1.5\times 284g/mol=426g$