1. Given reaction

$2Cu_2O + Cu_2S \rightarrow 6Cu + SO_2$

We first calculate molar masses.

2. Molar masses

For $Cu_2O$:

$M(Cu_2O)=2(63.5)+16=127+16=143\,\text{g mol}^{-1}$

For $Cu_2S$:

$M(Cu_2S)=2(63.5)+32=127+32=159\,\text{g mol}^{-1}$

3. Moles of reactants

Mass of $Cu_2O = 2.86\times 10^3 = 2860\,\text{g}$

$n(Cu_2O)=\frac{2860}{143}=20\,\text{mol}$

Mass of $Cu_2S = 4.77\times 10^3 = 4770\,\text{g}$

$n(Cu_2S)=\frac{4770}{159}=30\,\text{mol}$

4. Find the limiting reagent

From the balanced equation:

$2\,\text{mol } Cu_2O \text{ react with } 1\,\text{mol } Cu_2S$

For $20$ mol $Cu_2O$, required $Cu_2S$ is:

$\frac{20}{2}=10\,\text{mol}$

Available $Cu_2S = 30$ mol, so $Cu_2S$ is in excess.

Hence, $Cu_2O$ is the limiting reagent.

5. Moles of copper produced

From the reaction:

$2Cu_2O \rightarrow 6Cu$

So,

$1\,\text{mol } Cu_2O \rightarrow 3\,\text{mol } Cu$

Therefore, from $20$ mol $Cu_2O$:

$n(Cu)=20\times 3=60\,\text{mol}$

6. Mass of copper produced

$m(Cu)=60\times 63.5=3810\,\text{g}$

7. Final answer

The mass of copper produced is:

$\boxed{3810\,\text{g}}$