Calcium phosphide `Ca_(3)P_(2)` formed by reacting magnesium with excess calcium orthophosphate `Ca_(s)(PO_(4))_(2)` was hydrolysed by excess water. The evolved. Phosphine `PH_(5)` was burnt in air to yield phosphrous pentoxide `(P_(2)O_(6))` . How many gram of magnesium metaphosphate would be obtain if 192 gram Mg were used (Atomic weight of Mg=24, P=31)
`Ca_(3)(PO_(4))_(2) + Mg to Ca_(3)P_(2) +_ MgO`
`Ca_(3)P_(2)+H_(2)O to Ca(OH)_(2) + PH_(3)`
`PH_(3) + O_(2) to P_(2)O_(5) + H_(2)O`
`MgO + P_(2)O_(5) to Mg(PO_(3))_(2)`
`" "` magnesium metaphosphate.

Explanation

To solve the problem step by step, we will analyze the reactions and perform stoichiometric calculations to find the mass of magnesium metaphosphate (Mg(PO3)2) produced from 192 grams of magnesium (Mg).

### Step 1: Write down the relevant reactions
The reactions given in the problem are:
1. \( \text{Ca}_3(\text{PO}_4)_2 + \text{Mg} \rightarrow \text{Ca}_3\text{P}_2 + \text{MgO} \)
2. \( \text{Ca}_3\text{P}_2 + \text{H}_2\text{O} \rightarrow \text{Ca(OH)}_2 + \text{PH}_3 \)
3. \( \text{PH}_3 + \text{O}_2 \rightarrow \text{P}_2\text{O}_5 + \text{H}_2\text{O} \)
4. \( \text{MgO} + \text{P}_2\text{O}_5 \rightarrow \text{Mg(PO}_3)_2 \)
...