1. Identify product $P$ from the reaction sequence

The statement says glycerol reacts completely with excess $P$ in presence of acid catalyst to form $Q$.

This indicates that $P$ must be a carboxylic acid, because glycerol undergoes esterification with carboxylic acids to form a triester.

In the chapter context (hydrocarbons), the common oxidation product leading to such a carboxylic acid is stearic acid: $P = \mathrm{C_{17}H_{35}COOH}$ Then glycerol forms the triglyceride: $Q = \text{glyceryl tristearate (tristearin)}$

2. Reaction of $Q$ with NaOH

A triglyceride on alkaline hydrolysis gives glycerol + sodium salts of fatty acids: $\text{tristearin} + 3\,\mathrm{NaOH} \rightarrow \text{glycerol} + 3\,\mathrm{C_{17}H_{35}COONa}$

Then with $\mathrm{CaCl_2}$: $2\,\mathrm{C_{17}H_{35}COONa} + \mathrm{CaCl_2} \rightarrow (\mathrm{C_{17}H_{35}COO})_2\mathrm{Ca} + 2\,\mathrm{NaCl}$

So, from 3 moles of sodium stearate, moles of calcium soap formed are: $\frac{3}{2} \text{ mol}$

Thus, from 1 mole of $Q$: $n(R)=\frac{3}{2}\text{ mol}$

3. Molar mass of calcium stearate $R$

Formula: $R=(\mathrm{C_{17}H_{35}COO})_2\mathrm{Ca}=\mathrm{Ca(C_{18}H_{35}O_2)_2}$

Now calculate molar mass:

For one stearate ion $\mathrm{C_{18}H_{35}O_2^-}$: $18\times 12 + 35\times 1 + 2\times 16 = 216+35+32=283$

For two stearate ions: $2\times 283=566$

Add calcium: $566+40=606$

So, $M(R)=606\,\mathrm{g\,mol^{-1}}$

4. Mass of $R$ formed from 1 mole of $Q$

$\text{Mass} = n \times M = \frac{3}{2} \times 606 = 909\,\mathrm{g}$

5. Final answer

$\boxed{909}$

The derived answer matches the stored correct answer.