P2O5 on treatment with excess of H2o followed by excess of NH4oh from (Nh2HPO If 100 g of (NH2HPO4 is formed then find out the mass of P2O5 initially taken?
Answers
Molecular weight of P2O5 : 2*31+ 16*5 = 142
Molecular weight of (NH2) HPO4 : 14+3+31+64 = 112
100gms of NH2 HPO4 = 100/112 = 25/28 moles come from
=> 25/56 moles of P2 O5
=> 25/56 * 142 = 63.39 gms.
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Balanced equations:
P2 O5 + 3 H2 O => 2 H3 PO4
2 H3 PO4 (aq) + 2 NH4 OH (aq) ==> 2(NH2) HPO4 (s)+3H2 (g)+ 2H2O
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We know that in the final product, Phosphorus is present only in (NH2)HPO4. It contains one atom of P. There are two atoms of P in P2O5. Hence, we know easily that 1 mole of pentoxide gives (NH2)HPO4 (s) precipitate.
Answer:
Explanation:1 mole of P2O5 (Phosphorus Pentoxide) with excess water gives 2 moles of Phosphoric acid H3 PO4. 2 moles of Phosphoric acid in excess of Ammonium Hydroxide, produces 2 moles of (NH2) HPO4 salt (s) and Hydrogen gas.
Molecular weight of P2O5 : 2*31+ 16*5 = 142
Molecular weight of (NH2) HPO4 : 14+3+31+64 = 112
100gms of NH2 HPO4 = 100/112 = 25/28 moles come from
=> 25/56 moles of P2 O5
=> 25/56 * 142 = 63.39 gms.
====
Balanced equations:
P2 O5 + 3 H2 O => 2 H3 PO4
2 H3 PO4 (aq) + 2 NH4 OH (aq) ==> 2(NH2) HPO4 (s)+3H2 (g)+ 2H2O
============
We know that in the final product, Phosphorus is present only in (NH2)HPO4. It contains one atom of P. There are two atoms of P in P2O5. Hence, we know easily that 1 mole of pentoxide gives (NH2)HPO4 (s) precipitate.
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