Question

What mass of phosphoric acid (h3po4, 98.0 g/mol) is produced from the reaction of 10.0 g of p4o10 (284 g/mol) with excess water?

Answer 1

Answer : The mass of phosphoric acid produced is 6.89 grams.

Solution : Given,

Mass of
`P_4O_(10)` = 10.0 g

Molar mass of
`P_4O_(10)` = 284 g/mole

Molar mass of
`H_3PO_4` = 98.0 g/mole

First we have to calculate the moles of
`P_4O_(10)`.

`\text{ Moles of }P_4O_(10)=\frac{\text{ Mass of }P_4O_(10)}{\text{ Molar mass of }P_4O_(10)}=(10.0g)/(284g/mole)=0.0352moles`

Now we have to calculate the moles of
`MgO`

The balanced chemical reaction is,

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

From the reaction, we conclude that

As, 1 mole of
`P_4O_(10)` react to give 2 mole of
`H_3PO_4`

So, 0.0352 moles of
`P_4O_(10)` react to give
`0.0352* 2=0.0704` moles of
`H_3PO_4`

Now we have to calculate the mass of
`H_3PO_4`

`\text{ Mass of }H_3PO_4=\text{ Moles of }H_3PO_4* \text{ Molar mass of }H_3PO_4`

`\text{ Mass of }H_3PO_4=(0.0704moles)* (98.0g/mole)=6.89g`

Therefore, the mass of phosphoric acid produced is 6.89 grams.

Answer 2

when the balanced reaction equation is:
P4O10 + 6H2O→ 4 H3PO4
when we have the mass of P4O10 = 10 g and the molar mass of P4O10=284 g/mol & we have the molar mass of H3PO4 =98 g/mol so we can get the mass of H3PO4 by substitution by:
mass of H3PO4 = (mass of P4O10)/(molar mass of P4O10) * 4(mol of H3PO4)*molar mass of H3PO4
∴mass of H3PO4 = (10 / 284) * 4 * 98 = 13.8 g