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How many grams of `CaO` are required to neutralise `852 g` of `P_4 O_10` ? Draw the structure of `P_4 O_10`.A. 852gB. 1008gC. 85gD. 7095g

Answer» `6CaO+P_4O_10 to 2Ca_3 (PO_4)_2`
Mole of `P_4O_10` = molar mass of `P_4O_10` = 284g
852g of `P_4O_10=852/284=3` mol
mole of `P_4O_10` reacts with 6 moles of CaCO
3 moles of `P_4O_10` reacts with 18 moles of CaCO
Mass of 18 moles of CaO `=18xx56=1008g`


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