Exactly 4.0 g of sulfur is burnt in a fume extraction cupboard. The sulfur is ignited in a excess of oxygen: \[ \mathrm{S}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{SO}_{2}(\mathrm{~g}) \] a. What volume of gas will form in \( \mathrm{dm}^{3} \) at rtp?

To find the volume of gas that will form when 4.0 g of sulfur is burnt in excess oxygen, we need to follow these steps: 1. Calculate the number of moles of sulfur burnt. 2. Use the balanced chemical equation to determine the stoichiometry of the reaction. 3. Calculate the number of moles of gas produced. 4. Use the ideal gas law to calculate the volume of gas at RTP (room temperature and pressure). Let's start by calculating the number of moles of sulfur burnt: Given: - Mass of sulfur burnt = 4.0 g - Molar mass of sulfur (S) = 32.06 g/mol Number of moles of sulfur burnt = Mass of sulfur burnt / Molar mass of sulfur Let's calculate the number of moles of sulfur burnt. Calculate the value by following steps: - step0: Calculate: \(\frac{4}{32.06}\) - step1: Convert the expressions: \(\frac{4}{\frac{1603}{50}}\) - step2: Multiply by the reciprocal: \(4\times \frac{50}{1603}\) - step3: Multiply: \(\frac{4\times 50}{1603}\) - step4: Multiply: \(\frac{200}{1603}\) The number of moles of sulfur burnt is approximately 0.1248 mol. Next, we need to determine the stoichiometry of the reaction. From the balanced chemical equation: \[ \mathrm{S}(\mathrm{s}) + \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{SO}_{2}(\mathrm{g}) \] We can see that 1 mole of sulfur reacts to produce 1 mole of sulfur dioxide (SO₂). Therefore, the number of moles of sulfur dioxide produced is also 0.1248 mol. Now, we will use the ideal gas law to calculate the volume of gas at RTP. The ideal gas law is given by: \[ PV = nRT \] Where: - P is the pressure (1 atm at RTP) - V is the volume (to be calculated) - n is the number of moles of gas (0.1248 mol) - R is the ideal gas constant (0.0821 L·atm/mol·K) - T is the temperature in Kelvin (273.15 K at RTP) We can rearrange the ideal gas law to solve for volume: \[ V = \frac{nRT}{P} \] Let's substitute the values and calculate the volume of gas at RTP. Calculate the value by following steps: - step0: Calculate: \(\frac{\left(0.1248\times 0.0821\times 273.15\right)}{1}\) - step1: Remove the parentheses: \(\frac{0.1248\times 0.0821\times 273.15}{1}\) - step2: Multiply the terms: \(2.798716752\) The volume of gas that will form when 4.0 g of sulfur is burnt in excess oxygen at RTP is approximately 2.7987 dm³.