Statement I : When 124 g of  $P_4$ and 256 g  of  $S_8$  react according to the reaction : 
$4P_4 + 5S_8\to 4P_4{S}_{10}$, then  $S_8$ is the limiting reagent. (Atomic weights: P = 31,S = 32)
Statement II : Amount of  $P_4{S}_{10}$  formed depends on the amount of  $P_4$ consumed in the reaction. 
In the light of the above statements, choose the most appropriate answer from the options given below:

The reagent which is completely consumed in the reaction is called limiting 
reagent or limiting reagent is the reagent which decides the yield of the product. $S_8$ is the limiting reagent in the given reaction and the amount of $P_4{S}_{10 }$formed depends on the amount of $P_4$ used.
  $4P_4+5S_8\to 4P_4{S}_{10}$
$\frac{\text{124}}{\text{31}}\text{ = 4mol }\frac{\text{256}}{\text{256}}\text{ = 1mol}$ –
       $4-\frac{4}{5}=\frac{16}{5}$     –    $\frac{4}{5}mol$