for the reaction is:

$$\mathrm{NaCl} + \mathrm{AgNO}_3 \rightarrow \mathrm{AgCl} + \mathrm{NaNO}_3$$

From the equation, we can see that one mole of $$\mathrm{NaCl}$$ reacts with one mole of $$\mathrm{AgNO}_3$$ to produce one mole of $$\mathrm{AgCl}$$.

First, we need to find the number of moles of $$\mathrm{NaCl}$$ and $$\mathrm{AgNO}_3$$ present in the given masses.

The molar mass of $$\mathrm{NaCl}$$ is:

$$\mathrm{M_{NaCl}} = 23 + 35.5 = 58.5 \, \mathrm{g/mol}$$

So, the number of moles of $$\mathrm{NaCl}$$ in $$11.70 \, \mathrm{g}$$ is:

$$\mathrm{n_{NaCl}} = \frac{11.70 \, \mathrm{g}}{58.5 \, \mathrm{g/mol}} = 0.20 \, \mathrm{mol}$$

The molar mass of $$\mathrm{AgNO}_3$$ is:

$$\mathrm{M_{AgNO_3}} = 108 + 14 + (3 \times 16) = 108 + 14 + 48 = 170 \, \mathrm{g/mol}$$

So, the number of moles of $$\mathrm{AgNO}_3$$ in $$3.4 \, \mathrm{g}$$ is:

$$\mathrm{n_{AgNO_3}} = \frac{3.4 \, \mathrm{g}}{170 \, \mathrm{g/mol}} = 0.02 \, \mathrm{mol}$$

Since the ratio of $$\mathrm{NaCl}$$ to $$\mathrm{AgNO}_3$$ in the reaction is 1:1, the limiting reagent is $$\mathrm{AgNO}_3$$, because it has fewer moles (0.02 moles) compared to $$\mathrm{NaCl}$$ (0.20 moles).

Now, we calculate the mass of $$\mathrm{AgCl}$$ produced from 0.02 moles of $$\mathrm{AgNO}_3$$.

The molar mass of $$\mathrm{AgCl}$$ is:

$$\mathrm{M_{AgCl}} = 108 + 35.5 = 143.5 \, \mathrm{g/mol}$$

So, the mass of $$\mathrm{AgCl}$$ produced from 0.02 moles is:

$$\mathrm{m_{AgCl}} = 0.02 \, \mathrm{mol} \times 143.5 \, \mathrm{g/mol} = 2.87 \, \mathrm{g}$$

Therefore, the mass of $$\mathrm{AgCl}$$ precipitated is 2.87 g, making the correct answer:

Option C: 2.87 g

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