The mass of $P_4{O}_{10}$ produced, if $440$ g of $P_4{S}_3$ is mixed with $384$ g of $O_2$ is :
$P_4{S}_3+O_2⟶P_4{O}_{10}+{SO}_2$

The correct option is C 426 g
Balanced reaction:
$P_4{S}_3+8O_2\to P_4{O}_{10}+3S{O}_2$
Mass given: 440 g 384 g
No. of mole: 440/221 384/32
= 1.99 =12
Ratio of mole given 1.99/1 12/8
and stoichiometry =1.99 =1.5
coefficient :
As 1.5 is lower than 1.99 so, $O_2$ is limiting reagent and product will
form according to no. of mole of $O_2$ (given)
So, Mole produced of $P_4{O}_{10}=\frac{1}{8}\times 12=1.5mole$
And mass produced of $P_4{O}_{10}=1.5\times 284=426g$