1.

How many grams of CaO are required to neutralise 852g of P_(4)O_(10)?

Answer»

Solution :The REACTION will be:
`6CaO+P_(4)O_(10)rarr2Ca_(3)(PO_(4))_(2)`
`852g P_(4)O_(10) m3 "mol" P_(4)O_(10)`
1 MOLE of `P_(4)O_(10)` neutralises 6 MOLES of `CAO`
`therefore 3 "moles of" P_(4)O_(10)` will neutralise 18 moles of CaO.
`therefore "mass of" CaO=18xx56=1008g`


Discussion

No Comment Found