Question

P4O10 + 6H20 → 4H3PO4

If 30 g of H2O reacts with 40g of P4O10, what is the limiting reactant? How much H3PO4 can be formed?

Answer 1

Answer:
`P_4O_(10)` is the limiting reagent

66.6 g of
`H_3PO_4` will be formed.

Step-by-step explanation:

To calculate the moles :

`\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}`
`\text{Moles of} P_4O_(10)=(40g)/(284g/mol)=0.14moles`

`\text{Moles of} H_2O=(30g)/(18g/mol)=1.7moles`

`P_4O_(10)+6H_2O\rightarrow 4H_3PO_4`

According to stoichiometry :

1 mole of
`P_4O_(10)` require 6 moles of
`H_2O`

Thus 0.14 moles of
`P_4O_(10)` will require=
`(6)/(1)* 0.14=0.84moles` of
`H_2O`

Thus
`P_4O_(10)` is the limiting reagent as it limits the formation of product and
`H_2O` is the excess reagent.

As 1 mole of
`P_4O_(10)` give = 4 moles of
`H_3PO_4`

Thus 0.17 moles of
`P_4O_(10)` give =
`(4)/(1)* 0.17=0.68moles` of
`H_3PO_4`

Mass of
`H_3PO_4=moles* {\text {Molar mass}}=0.68moles* 98g/mol=66.6g`

Thus 66.6 g of
`H_3PO_4` will be formed.