Answer:  [tex]P_4O_{10}[/tex] is the limiting reagent

66.6 g of [tex]H_3PO_4[/tex] will be formed.

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]  [tex]\text{Moles of} P_4O_{10}=\frac{40g}{284g/mol}=0.14moles[/tex]

[tex]\text{Moles of} H_2O=\frac{30g}{18g/mol}=1.7moles[/tex]

[tex]P_4O_{10}+6H_2O\rightarrow 4H_3PO_4[/tex]  

According to stoichiometry :

1 mole of [tex]P_4O_{10}[/tex] require 6 moles of [tex]H_2O[/tex]

Thus 0.14 moles of [tex]P_4O_{10}[/tex] will require=[tex]\frac{6}{1}\times 0.14=0.84moles[/tex]  of [tex]H_2O[/tex]

Thus [tex]P_4O_{10}[/tex] is the limiting reagent as it limits the formation of product and [tex]H_2O[/tex] is the excess reagent.

As 1 mole of [tex]P_4O_{10}[/tex] give = 4 moles of [tex]H_3PO_4[/tex]

Thus 0.17 moles of [tex]P_4O_{10}[/tex] give =[tex]\frac{4}{1}\times 0.17=0.68moles[/tex]  of [tex]H_3PO_4[/tex]

Mass of [tex]H_3PO_4=moles\times {\text {Molar mass}}=0.68moles\times 98g/mol=66.6g[/tex]

Thus 66.6 g of [tex]H_3PO_4[/tex] will be formed.