How many grams of CaO are required to neutralise 852 g of $P_{4} O_{10}$ ?

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Solution

The reaction will be:
$6 C a O + P_{4} O_{10} \to 2 C a_{3} (P O_{4})_{2}$
852 g $P_{4} O_{10} \equiv 3 m o l P_{4} O_{10}$
1 mole of $P_{4} O_{10}$ will neutralises 6 moles of CaO.
$∴$ 3 moles of $P_{4} O_{10}$ will neutralise 18 moles of CaO.
$∴$ Mass of CaO $= 18 \times 56 = 1008 g$

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