Phosphoric acid `(H_3PO_4)` prepared in a two step process. (1) `P_4+5O_2rarrP_4 O_10` (2) `P_4 O_10+6H_2Orarr4H_3 PO_4` We allow `62g` of phosphorus to react with excess oxygen which form `P_4 O_10` in `85%` yield. In the step (2) reaction `90%` yield of H_3PO_4` is obtained. mass of `H_3PO_4` produced is :

Phospheric acid (H_(3)PO_(4)) perpared in two step process . (1) P_(4)+5O_(2) to P_(4)O_(10)" "(2) P_(4)O_(10) + 6H_(2)O to 4H_(3)PO_(4) Well allow 62 g of phosphrous to react with exces oxygen which from P_(4)O_(10) in 85% yield . In the sep (2) reaction 90% yield of H_3)PO_(4) is obtained . Mass of H_(3) PO_(4) produced is :

In the reaction, H_3PO_4+Ca(OH)_2rarrCaHPO_4+2H_2O, the equivalent mass of H_3PO_4 is

The equivalent weight of phosphoric acid (H_(3)PO_(4)) in the reaction NaOH+H_(3)PO_(4) rarr NaH_(2)PO_(4)+H_(2)O is

How many moles of H_3 PO_4 are obtained on hydrolysis of one mole of P_4 O_8 .

The true statement of the acids of phosphorus H_3 PO_2, H_3 PO_2 and H_3 PO_4 is

What is the equivalent weight of phosphoric acid (H_(3) PO_(4)) according to the equation NaOH + H_(3) PO_(4) rarr NaH_(2) PO_(4) + H_(2)O

In the reaction P_4 + 3 KOH + 3H_2 O to PH_3 + 3KH_2 PO_2