The percentage of P_2O_5 in diammonium hydrog | vedclass

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(C) The molar mass of $(NH_4)_2HPO_4$ is $2 	imes (14 + 4) + 1 + 31 + 4 	imes 16 = 36 + 1 + 31 + 64 = 132 \ g/mol$. Since $2$ moles of $(NH_4)_2HPO_

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$53.78$

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(C) The molar mass of $(NH_4)_2HPO_4$ is $2 	imes (14 + 4) + 1 + 31 + 4 	imes 16 = 36 + 1 + 31 + 64 = 132 \ g/mol$. Since $2$ moles of $(NH_4)_2HPO_4$ contain $1$ mole of $P_2O_5$,the mass of $2$ moles of $(NH_4)_2HPO_4$ is $2 	imes 132 = 264 \ g$. The molar mass of $P_2O_5$ is $2 	imes 31 + 5 	imes 16 = 62 + 80 = 142 \ g/mol$. Percentage of $P_2O_5 = \frac{	ext{Mass of } P_2O_5}{	ext{Mass of } 2 	ext{ moles of } (NH_4)_2HPO_4} 	imes 100$. Percentage of $P_2O_5 = \frac{142}{264} 	imes 100 = 53.78\%$.

The percentage of $P_2O_5$ in diammonium hydrogen phosphate $(NH_4)_2HPO_4$ is

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