Sulfur dioxide will react with water to form sulfurous acid : SO2(g) + H2O(l) → H2SO3(l) What mass of sulfur dioxide is needed to prepare 27.86 g of H2SO3(l)?
28.77
42.24g
how many moles of H2SO3 in 27.86 grams?
The equation says you will need that many moles of SO2. Convert that to grams.
To determine the mass of sulfur dioxide needed to prepare a certain amount of sulfurous acid (H2SO3), we'll use the balanced chemical equation to establish the ratio of sulfur dioxide to sulfurous acid.
1. Begin by calculating the molar mass of H2SO3 (sulfurous acid):
H2SO3 = (2 * H) + S + (3 * O)
= (2 * 1.01 g/mol) + 32.07 g/mol + (3 * 16.00 g/mol)
= 62.09 g/mol
2. Next, convert the given mass of H2SO3 to moles using the molar mass calculated above:
Moles of H2SO3 = Mass of H2SO3 / Molar mass of H2SO3
= 27.86 g / 62.09 g/mol
= 0.4487 mol
3. Now, refer to the balanced chemical equation:
SO2(g) + H2O(l) → H2SO3(l)
From the equation, we can determine the stoichiometry of sulfur dioxide to sulfurous acid:
1 mol SO2 produces 1 mol H2SO3
4. Apply the stoichiometric ratio from step 3 to calculate the moles of sulfur dioxide required:
Moles of SO2 = Moles of H2SO3
= 0.4487 mol
5. Finally, calculate the mass of sulfur dioxide using the molar mass of SO2 (32.07 g/mol):
Mass of SO2 = Moles of SO2 * Molar mass of SO2
= 0.4487 mol * 32.07 g/mol
≈ 14.402 g
Therefore, approximately 14.402 grams of sulfur dioxide is needed to prepare 27.86 grams of sulfurous acid.