Question

A \(1.500-\mathrm{g}\) sample containing sodium nitrate was heated to form \(\mathrm{NaNO}_{2}\) and \(\mathrm{O}_{2}\). The oxygen evolved was collected over water at \(23^{\circ} \mathrm{C}\) and \(752 \mathrm{~mm} \mathrm{Hg} ;\) its volume was \(125.0 \mathrm{~mL} .\) Calculate the percentage of \(\mathrm{NaNO}_{3}\) in the sample. The vapor pressure of water at \(23^{\circ} \mathrm{C}\) is \(21.07 \mathrm{~mm} \mathrm{Hg}\).

Short answer

Answer: The percentage of sodium nitrate (NaNO3) in the 1.500 g sample is 58.73%.

Step-by-step solution

Convert the volume of collected oxygen gas to moles

First, we need to convert the given pressure to account for the vapor pressure of water. Given that the vapor pressure of water at 23°C is 21.07 mm Hg, we subtract it to find the partial pressure of oxygen: Partial pressure of O2 = 752 mm Hg - 21.07 mm Hg = 730.93 mm Hg Now, we can use the Ideal Gas Law to find the moles of O2. Recall that the Ideal Gas Law is given as: PV = nRT where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin. We have P = 730.93 mm Hg, V = 125.0 mL, R = 62.363 L·mm Hg/(K·mol), and T = 23°C + 273.15 = 296.15 K. We can now solve for n: n(O2) = (P * V) / (R * T) = (730.93 mm Hg * 125.0 mL) / (62.363 L·mm Hg/(K·mol) * 296.15 K) n(O2) = 0.00518 moles of O2

Calculate the moles of NaNO3 responsible for generating the oxygen

The balanced chemical equation for the reaction is: 2NaNO3 → 2NaNO2 + O2 From the stoichiometry of the reaction, we can see that for each mole of O2 produced, 2 moles of NaNO3 are consumed. Thus, we can find the moles of NaNO3 responsible for generating the collected oxygen: n(NaNO3) = 2 * n(O2) = 2 * 0.00518 moles = 0.01036 moles of NaNO3

Convert moles of NaNO3 to grams and calculate the percentage of NaNO3 in the sample

Next, we need to convert the moles of NaNO3 found in step 2 to grams. To do this, we multiply the moles of NaNO3 by its molecular weight: Mass of NaNO3 = n(NaNO3) * (Molar mass of NaNO3) = 0.01036 moles * 85.0 g/mol = 0.881 g Finally, we can find the percentage of NaNO3 in the sample by dividing the mass of NaNO3 by the total mass of the sample and multiplying by 100: Percentage of NaNO3 = (Mass of NaNO3 / Mass of sample) * 100 = (0.881 g / 1.500 g) * 100 = 58.73% Therefore, the percentage of NaNO3 in the sample is 58.73%.