Question

How many grams of `CaO` are required to neutralise `852 g` of `P_4 O_10` ? Draw the structure of `P_4 O_10`.

Answer 1

For structure of `P_4 O_10`,
`P_4 O_10 + 6 CaO rarr 2 Ca_3(PO_4)_2`
`6 CaO -= P_4 O_10`
`("Weight of" CaO)/(Mw of CaO xx 6) = ("Weight of" P_4 O_10)/(Mw of P_4 O_10)`
`(w)/(6(40 + 16)) = (852)/((31 xx 4)(16 + 10))`
`(w)/(336) = (852)/(284)`
`w = 1008 of CaO`.