A mass of 6.35 g of pure copper was reacted with an excess of warm concentrated sulfuric acid to produce copper(II) sulfate, water and sulfur dioxide according to the following equation: Cu + 2H₂SO₄ → CuSO₄ + 2H₂O + SO₂ - Leaving Cert Chemistry - Question a - 2010

From the balanced equation, 2 moles of water (H₂O) are produced for every 1 mole of copper reacted. Thus:

$n_{H₂O} = 2 \times n_{Cu} = 2 \times 0.1 = 0.2 \text{ moles}$

Now, to find the mass of water produced, we calculate:

$m_{H₂O} = n_{H₂O} \times M_{H₂O}$ where the molar mass of water (H₂O) is approximately 18 g/mol:

$m_{H₂O} = 0.2 \text{ moles} \times 18 \text{ g/mol} = 3.6 ext{ g}$

From the balanced equation, 1 mole of sulfur dioxide (SO₂) is produced for every mole of copper reacted. Therefore:

$n_{SO₂} = n_{Cu} = 0.1 \text{ moles}$

The volume of gas at standard temperature and pressure (s.t.p.) is measured as 22.4 liters per mole:

$V_{SO₂} = n_{SO₂} \times 22.4 \text{ L/mol} = 0.1 \text{ moles} \times 22.4 \text{ L/mol} = 2.24 \text{ liters}$

To find the number of molecules in this volume, we use Avogadro's number ($6.022 \times 10^{23}$ molecules/mol):

$\text{Number of molecules} = n_{SO₂} \times 6.022 \times 10^{23} = 0.1 \text{ moles} \times 6.022 \times 10^{23} \approx 6 \times 10^{22} \text{ molecules}$