Calculate the mass of copper produced from 5.0 tonnes of copper carbonate, given the reaction 2CuCO3 + C -> 2Cu + 3CO2.

We'll use the following relative atomic masses ($\text{A}_r$): Cu = 63.5, C = 12.0, O = 16.0.
Molar mass of copper carbonate ($\text{CuCO}_3$):

$$M(\text{CuCO}_3) = \text{A}_r(\text{Cu}) + \text{A}_r(\text{C}) + (3 \times \text{A}_r(\text{O}))$$ $$M(\text{CuCO}_3) = 63.5 + 12.0 + (3 \times 16.0) = 63.5 + 12.0 + 48.0 = 123.5 \text{ g/mol}$$

Molar mass of copper ($\text{Cu}$):

$$M(\text{Cu}) = \text{A}_r(\text{Cu}) = 63.5 \text{ g/mol}$$

Given mass of copper carbonate = 5.0 tonnes. Since 1 tonne = $10^6$ g, we have $5.0 \times 10^6 \text{ g}$.

$$\text{Moles of CuCO}_3 = \frac{\text{Mass of CuCO}_3}{\text{Molar mass of CuCO}_3}$$ $$\text{Moles of CuCO}_3 = \frac{5.0 \times 10^6 \text{ g}}{123.5 \text{ g/mol}}$$

The balanced equation is $2\text{CuCO}_3 + \text{C} \rightarrow 2\text{Cu} + 3\text{CO}_2$.
From the equation, 2 moles of $\text{CuCO}_3$ produce 2 moles of $\text{Cu}$. This means the mole ratio is 1:1.
Therefore, moles of $\text{Cu}$ produced = moles of $\text{CuCO}_3$.

$$\text{Moles of Cu} = \frac{5.0 \times 10^6}{123.5} \text{ mol}$$

$$\text{Mass of Cu} = \text{Moles of Cu} \times \text{Molar mass of Cu}$$ $$\text{Mass of Cu} = \left(\frac{5.0 \times 10^6}{123.5} \text{ mol}\right) \times (63.5 \text{ g/mol})$$ $$\text{Mass of Cu} = \frac{5.0 \times 10^6 \times 63.5}{123.5} \text{ g}$$ $$\text{Mass of Cu} = \frac{317.5 \times 10^6}{123.5} \text{ g}$$ $$\text{Mass of Cu} \approx 2.57085 \times 10^6 \text{ g}$$

$$\text{Mass of Cu in tonnes} = \frac{2.57085 \times 10^6 \text{ g}}{10^6 \text{ g/tonne}}$$ $$\text{Mass of Cu in tonnes} \approx 2.57085 \text{ tonnes}$$

Rounding to two significant figures (as per the input 5.0 tonnes):

$$\text{Mass of Cu} = \boxed{\text{2.6 tonnes}}$$