When 5.00 mL of a sodium nitrite solution containing 17.04 g/L of sodium nitrite

Question

When 5.00 mL of a sodium nitrite solution containing 17.04 g/L of sodium nitrite is added to 25.00 mL of sulfamic acid solution containing 2.754 g/L of sulfamic acid, the following reaction occurs:

HSO3NH2 (aq) + NaNO2 (aq) --> N2(g) + NaHSO4 (aq) + H2O (l)

a) Determie the mass of nitrogen gas produced, and calculate the volume occupied by this mass of gas at a temperature of 23 degrees C and a pressure of 00.5 kPa

b) When this experiment was carried out by a student and the nitrogen gas was collected over water, the volume of “wet” nitrogen produced (at 23°C and 99.5 kPa) was 14.81 mL. What was the percentage yield obtained by the student? The vapour pressure of water at 23°C is 2.81 kPa.

Explanation / Answer

For the given reaction,

a) moles of NaNO2 taken = [(17.04 g/L)/69 g/mol)] x 0.005 L = 0.00123 mol

moles HSO3NH2 taken = [(2.754 g/L)/97.1 g/mol] x 0.025 L = 0.00071 mol (Theoretical yield)

since moles of HSO3NH2 is lower, this is the limiting reactant

mass of N2 produced = 0.00071 mol x 28 g/mol = 0.02 g

Using ideal gas relation,

Volume (V) = nRT/P

n = 0.00071 mol

R = gas constant

T = 23 oC + 273 = 296 K

P = 0.5 kPa = 0.005 atm

so,

Volume occupied by N2 = 0.00071 x 0.08205 x 296/0.005 = 3.45 L

b) Volume of wet N2 produced in experiment = 14.81 ml = 0.01481 L

Pressure of dry N2 = 99.5 kPa - 2.81 kPa = 96.69 kPa = 0.954 atm

Temperature (T) = 23 + 273 = 296 K

moles N2 produced (n) = PV/RT

                                       = 0.954 x 0.01481/0.08205 x 296 = 0.000582 mol (actual yield of N2)

percentage yield = actual yield x 100/theoretical yield

                           = 0.000582 mol x 100/0.00071 mol

                           = 82%

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