How much copper(II) oxide (CuO) will react with 4.9 g of tetraoxosulphate(VI) acid (H2SO4)?

The reaction between tetraoxosulphate(VI) acid ($\text{H}_2\text{SO}_4$) and copper(II) oxide ($\text{CuO}$) is:

$$\text{H}_2\text{SO}_4 + \text{CuO} \rightarrow \text{CuSO}_4 + \text{H}_2\text{O}$$

This equation is already balanced, showing a $1:1$ mole ratio between $\text{H}_2\text{SO}_4$ and $\text{CuO}$.

Given atomic masses: $\text{Cu} = 64$, $\text{O} = 16$, $\text{S} = 32$, $\text{H} = 1$.

Molar mass of $\text{H}_2\text{SO}_4$:

$$(2 \times 1) + 32 + (4 \times 16) = 2 + 32 + 64 = 98 \, \text{g mol}^{-1}$$

Molar mass of $\text{CuO}$:

$$64 + 16 = 80 \, \text{g mol}^{-1}$$

From the balanced equation, 1 mole of $\text{H}_2\text{SO}_4$ reacts with 1 mole of $\text{CuO}$.
This means $98 \, \text{g}$ of $\text{H}_2\text{SO}_4$ reacts with $80 \, \text{g}$ of $\text{CuO}$.

We can set up a proportion to find the mass of $\text{CuO}$ ($x$) that reacts with $4.9 \, \text{g}$ of $\text{H}_2\text{SO}_4$:

$$\frac{\text{Mass of } \text{H}_2\text{SO}_4}{\text{Molar mass of } \text{H}_2\text{SO}_4} = \frac{\text{Mass of } \text{CuO}}{\text{Molar mass of } \text{CuO}}$$ $$\frac{4.9 \, \text{g}}{98 \, \text{g mol}^{-1}} = \frac{x \, \text{g}}{80 \, \text{g mol}^{-1}}$$

Now, solve for $x$:

$$x = \frac{4.9 \, \text{g}}{98 \, \text{g mol}^{-1}} \times 80 \, \text{g mol}^{-1}$$ $$x = 0.05 \, \text{mol} \times 80 \, \text{g mol}^{-1}$$ $$x = 4.0 \, \text{g}$$

The amount of copper(II) oxide that will react is $4.0 \, \text{g}$.

The final answer is $\text{D}$.