Question

DEAR SIR
IN DETERMINATION PERCENT Na3PO4 AS P2O5% WHAT IS EQUIVALENT GRAM P2O5,E=M/n WHAT IS n FOR P2O5?
BEST REGARDS

Answer 1

Answer on Question #52792 – Chemistry – Other

Question:

In determination percent $\mathrm{Na}_3\mathrm{PO}_4$ as $\mathrm{P}_2\mathrm{O}_5\%$ what is equivalent gram $\mathrm{P}_2\mathrm{O}_5$, $E = m/n$ what is $n$ for $\mathrm{P}_2\mathrm{O}_5$?

Answer:

$$\mathrm{P}_2\mathrm{O}_5 + 3\mathrm{Na}_2\mathrm{O} \rightarrow 2\mathrm{Na}_3\mathrm{PO}_4$$

The molecular weights for $\mathrm{Na}_3\mathrm{PO}_4$ and $\mathrm{P}_2\mathrm{O}_5$ are 164 and $142\ \mathrm{g/mol}$, respectively.

Thus, the mass fraction for $\mathrm{P}_2\mathrm{O}_5\ \mathrm{w}(\%) = [M_w(P_2O_5)/M_w(Na_3PO_4)] \times 100\% = 86.59\%$

The mass fraction for $\mathrm{Na}_3\mathrm{PO}_4$ from $\mathrm{P}_2\mathrm{O}_5\%$ is: $\mathrm{w}(\%) = [M_w(Na_3PO_4)/M_w(P_2O_5)] \times 100\% = 100/\mathrm{w}(P_2O_5) = 115.49\%$

The equivalent gram for $\mathrm{P}_2\mathrm{O}_5$ is $M_w/2$ which equals $142\ \mathrm{g\ mol^{-1}/2\ eq\ mol^{-1}} = 71\ \mathrm{g/eq}$

An equivalent is of 2 (n = 2), because the coefficient at the product (Na₃PO₄) in the reaction equals 2. It means that the one molecule of $\mathrm{P}_2\mathrm{O}_5$ gives 2 molecules $\mathrm{Na}_3\mathrm{PO}_4$ or a half of $\mathrm{P}_2\mathrm{O}_5$ forms 1 molecule of $\mathrm{Na}_3\mathrm{PO}_4$.

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