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

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

Answer 2

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 $P_4O_{10}$.

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

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 $0.0352\times 2=0.0704$ moles of $H_3PO_4$

Now we have to calculate the mass of $H_3PO_4$

$\text{ Mass of }H_3PO_4=\text{ Moles of }H_3PO_4\times \text{ Molar mass of }H_3PO_4$

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

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