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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?

Respuesta :

superman1987
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 : The mass of phosphoric acid produced is 6.89 grams.

Solution : Given,

Mass of [tex]P_4O_{10}[/tex] = 10.0 g

Molar mass of [tex]P_4O_{10}[/tex] = 284 g/mole

Molar mass of [tex]H_3PO_4[/tex] = 98.0 g/mole

First we have to calculate the moles of [tex]P_4O_{10}[/tex].

[tex]\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[/tex]

Now we have to calculate the moles of [tex]MgO[/tex]

The balanced chemical reaction is,

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

From the reaction, we conclude that

As, 1 mole of [tex]P_4O_{10}[/tex] react to give 2 mole of [tex]H_3PO_4[/tex]

So, 0.0352 moles of [tex]P_4O_{10}[/tex] react to give [tex]0.0352\times 2=0.0704[/tex] moles of [tex]H_3PO_4[/tex]

Now we have to calculate the mass of [tex]H_3PO_4[/tex]

[tex]\text{ Mass of }H_3PO_4=\text{ Moles of }H_3PO_4\times \text{ Molar mass of }H_3PO_4[/tex]

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

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

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