A 200 cm3 solution of 0.20 M silver nitrate was added to 100 cm3 of a 0.10 M metal chloride solution. Precipitation occurred. What is the formula of the metal chloride?

The volume of silver nitrate solution is $200 \text{ cm}^3 = 0.200 \text{ L}$ and its concentration is $0.20 \text{ M}$.

$$\text{Moles of AgNO}_3 = \text{Concentration} \times \text{Volume}$$ $$\text{Moles of AgNO}_3 = 0.20 \text{ mol/L} \times 0.200 \text{ L} = 0.040 \text{ mol}$$

The precipitation reaction is $\text{Ag}^+(aq) + \text{Cl}^-(aq) \rightarrow \text{AgCl}(s)$.
From the stoichiometry, 1 mole of $\text{Ag}^+$ reacts with 1 mole of $\text{Cl}^-$. Since $\text{AgNO}_3$ provides $\text{Ag}^+$ ions in a $1:1$ ratio, the moles of $\text{Ag}^+$ are equal to the moles of $\text{AgNO}_3$.

$$\text{Moles of Cl}^- = \text{Moles of Ag}^+ = \text{Moles of AgNO}_3 = 0.040 \text{ mol}$$

The volume of the metal chloride solution is $100 \text{ cm}^3 = 0.100 \text{ L}$ and its concentration is $0.10 \text{ M}$.

$$\text{Moles of metal chloride} = \text{Concentration} \times \text{Volume}$$ $$\text{Moles of metal chloride} = 0.10 \text{ mol/L} \times 0.100 \text{ L} = 0.010 \text{ mol}$$

Let the formula of the metal chloride be $\text{MCl}_x$. This means that 1 mole of $\text{MCl}_x$ contains $x$ moles of $\text{Cl}^-$.
We can find $x$ by taking the ratio of moles of chloride ions to moles of metal chloride.

$$x = \frac{\text{Moles of Cl}^-}{\text{Moles of metal chloride}} = \frac{0.040 \text{ mol}}{0.010 \text{ mol}} = 4$$

Therefore, the formula of the metal chloride is $\text{MCl}_4$.

The correct option is A.
The right formula for the metal chloride is $\boxed{\text{A. MCl}_4}$.