Nitrogen and sulfur — practice questions

Step 1: Write the balanced equation for the reaction. $2NO(g) + O_2(g) \rightarrow 2NO_2(g)$

Step 2: Use the molar ratios from the equation. According to Avogadro's law, for gases at the same temperature and pressure, the ratio of volumes is equal to the ratio of moles. The ratio of NO to $NO_2$ is 2:2, which simplifies to 1:1.

Step 3: Calculate the volume of $NO_2$. Since the ratio is 1:1, the volume of $NO_2$ produced will be equal to the volume of NO reacted. Volume of $NO_2$ = Volume of NO = 120 cm³.

Answer: 120 cm³ of nitrogen dioxide can be formed.

Step 1: Calculate the moles of dichromate(VI) ions reacted. Moles = concentration × volume Moles of $Cr_2O_7^{2-}$ = 0.0200 mol dm⁻³ × (25.0 / 1000) dm³ = 5.00 × 10⁻⁴ mol.

Step 2: Use the molar ratio from the equation to find moles of $SO_2$. The ratio of $Cr_2O_7^{2-}$ to $SO_2$ is 1:3. Moles of $SO_2$ = 3 × Moles of $Cr_2O_7^{2-}$ = 3 × (5.00 × 10⁻⁴) = 1.50 × 10⁻³ mol.

Step 3: Calculate the mass of $SO_2$. Mass = moles × Molar mass ($M_r$) The $M_r$ of $SO_2$ = 32.1 + (2 × 16.0) = 64.1 g mol⁻¹. Mass of $SO_2$ = (1.50 × 10⁻³) mol × 64.1 g mol⁻¹ = 0.09615 g.

Answer: 0.0962 g (to 3 significant figures).