1.

How many grams of CaO are required to neutralise 852 g of `P_(4)O_(10)` ? Draw structure of `P_(4)O_(10)` molecule.

Answer» `underset(=336g)underset(6xx56)(6CaO) " " + " " underset(=284g)underset(4xx31+16xx10)(P_(4)O_(10)) " " to 2Ca_(3)(PO_(4))_(2)`
284 g of `P_(4)O_(10)` need CaO=336 g
852 g of `P_(4)O_(10)` need `CaO=((336g)xx(852g))/((284))=1008 g`
For structure of `P_(4)O_(10)`, consult Fig. 3 (Section 4)


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