The stoichiometric reaction of 516 g of dimethyldichlorosilane with water results in a tetrameric cyclic product X in 75% yield. The weight (in g) of X obtained is ____.

[Use, molar mass (g mol$$^{-1}$$): H = 1, C = 12, O = 16, Si = 28, Cl = 35.5]

Molecular formula of dimethyldichlorosilane: $$(CH_3)_2SiCl_2$$.

Molar mass of $$(CH_3)_2SiCl_2$$:
C: $$2 \times 12 = 24 \text{ g}$$
H: $$6 \times 1 = 6 \text{ g}$$
Si: $$1 \times 28 = 28 \text{ g}$$
Cl: $$2 \times 35.5 = 71 \text{ g}$$
Total $$= 24 + 6 + 28 + 71 = 129 \text{ g mol}^{-1}$$.

Moles of dimethyldichlorosilane taken:
$$n = \frac{516 \text{ g}}{129 \text{ g mol}^{-1}} = 4.0 \text{ mol}$$.

Hydrolysis-condensation stoichiometry to give the cyclic tetramer $$X$$ (octamethylcyclotetrasiloxane):
$$4\,(CH_3)_2SiCl_2 + 4\,H_2O \rightarrow [(CH_3)_2SiO]_4 + 8\,HCl$$.

Thus $$4$$ moles of starting material produce $$1$$ mole of $$X$$.
Therefore, theoretical moles of $$X$$ formed = $$\frac{4.0}{4} = 1.0 \text{ mol}$$.

Molar mass of $$X = [(CH_3)_2SiO]_4$$:
One $$(CH_3)_2SiO$$ unit: C$$_2$$H$$_6$$SiO $$= 30 + 28 + 16 = 74 \text{ g mol}^{-1}$$.
Tetramer (4 units): $$4 \times 74 = 296 \text{ g mol}^{-1}$$.

Theoretical mass of $$X$$ = $$1.0 \text{ mol} \times 296 \text{ g mol}^{-1} = 296 \text{ g}$$.

Given yield = $$75\%$$, actual mass obtained:
$$m = 0.75 \times 296 \text{ g} = 222 \text{ g}$$.

Hence, the weight of the tetrameric cyclic product $$X$$ obtained is 222 g.