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