A `5.0 g` quantity of white phosphorus was burned in an excess of oxygen and the product was dissolved in water to make `250.0 mL` solution.
(a) Write balanced equations for the reaction.
(b) When the solution was treated with an excess of aqueous `Ca(NO_(3))_(2)`, a white precipitate was obtained. What was it and how much it formed?
(c ) The precipitate in part (b) was removed and the solution was treated with an excess of `Zn`, yielding a colourless gas collected at `20^(@)C` and `742 mm of Hg`. What was the gas and how much volume of it was formed?

Explanation

(a) `P_(4)+5O_(2)rarrP_(4)O_(10)`
`{:("Moles at"t=0,(5)/(124),0,,),("Final moles",-,(5)/(124),,):}`
`P_(4)O_(10)+6H_(2)Orarr4H_(3)PO_(4)`
`:.` Moles of `H_(3)PO_(4)` formed `= 4xx"mole of"P_(4)O_(10)`
`=4xx(5)/(124)= 0.16`
(b) `3Ca(NO_(3))_(2)+2H_(3)PO_(4)rarrCa_(3)(PO_(4))_(2)+6HNO_(3)`
`:.` Moles of `HNO_(3)` formed `= 0.16xx(6)/(2)= 0.48`
(c ) `10HNO_(3)+4Znrarr4Zn(NO_(3))_(2)+underset("Colourless gas")(N_(2)O+5H_(2)O)`
`:.` Moles of `N_(2)O` formed `=(1)/(10)xx"mole of"HNO_(3)`
`=0.48xx(1)/(10)= 0.048`
`:. V=(nRT)/(P)=(0.048xx0.0821xx293xx760)/(742)`
`1.18"litre"`