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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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