Question

When aluminum metal is added to an aqueous solution of hydrochloric acid and ammonium perchlorate, an explosion results. The products of the explosion are aluminum chloride solid, nitrogen gas, and water vapor. The reaction is: 24 HCl(aq) + 10 Al(s) + 6 NH4CIO4(aq) → 3 N2(g) + 24 H₂O(g) + 10 AlCl3(s)

In an experiment, a 55.0-g sample of aluminum was combined with an excess of hydrochloric acid and ammonium perchlorate in aqueous solution. The resulting gasses produced from the explosion were collected at 127°C in a steel container with a volume of 67 L.

(a) Calculate the theoretical number of moles of nitrogen gas produced from the reaction.(b) Based on your answer in part (a), calculate the expected partial pressure of nitrogen gas in the steel container.

(c) Calculate the expected total pressure in the steel container.

(d) If the actual total pressure was found to be 2.43 atm, what was the percent yield of products from this reaction?​

Answer 1

The theoretical number of moles of nitrogen gas produced from the reaction is calculated using stoichiometry, ideal gas law is used to calculate the expected partial pressure and total pressure, and the percent yield of products is determined by comparing the actual yield of nitrogen gas to the theoretical yield.

Step-by-step explanation:

(a) To calculate the theoretical number of moles of nitrogen gas produced from the reaction, we can use the stoichiometry of the balanced equation. From the equation, we can see that for every 3 moles of N2 produced, we need 6 moles of NH4ClO4. We also know that the reaction produced a total of 55.0 g of aluminum, which can be converted to moles using the molar mass of aluminum. So, we can set up a proportion:

(10 mol Al / 26.98 g) = (3 mol N2 / x)

Solving for x, x = (3 mol N2 * 26.98 g) / 10 mol Al = 8.094 g N2

So, the theoretical number of moles of nitrogen gas produced is 8.094 g N2.

(b) To calculate the expected partial pressure of nitrogen gas in the steel container, we can use the ideal gas law. The equation is:

PV = nRT

Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. We know the volume is 67 L and the number of moles of nitrogen gas is 8.094 g / (28.02 g/mol) = 0.289 mol. The ideal gas constant R is 0.0821 L·atm/(mol·K). The temperature T is 127°C, which is 400.15 K. Plugging in these values, we can solve for P:

P = (0.289 mol * 0.0821 L·atm/(mol·K) * 400.15 K) / 67 L = 1.365 atm

So, the expected partial pressure of nitrogen gas in the steel container is 1.365 atm.

(c) To calculate the expected total pressure in the steel container, we need to take into account the partial pressure of nitrogen gas and the partial pressure of the other gases (water vapor and any remaining gases from hydrochloric acid and ammonium perchlorate). Since the reaction is balanced, we know that for every 3 moles of N2 produced, 24 moles of H2O vapor are produced. So, the mole ratio of nitrogen gas to water vapor is 3:24. We can use this ratio to calculate the moles of water vapor:

(0.289 mol N2 * 24 mol H2O) / 3 mol N2 = 2.312 mol H2O

Assuming all other gases from hydrochloric acid and ammonium perchlorate are at an equal pressure, the total number of moles of gases in the container is 0.289 mol N2 + 2.312 mol H2O = 2.601 mol. Plugging this value, along with the volume and temperature, into the ideal gas law equation:

P = (2.601 mol * 0.0821 L·atm/(mol·K) * 400.15 K) / 67 L = 1.549 atm

So, the expected total pressure in the steel container is 1.549 atm.

(d) To calculate the percent yield of products from this reaction, we need to compare the actual yield of nitrogen gas to the theoretical yield. The theoretical yield of nitrogen gas is 8.094 g, as calculated in part (a). The actual yield is not given in the question, so we need to calculate it. We know that the total pressure in the steel container is 2.43 atm. Using the ideal gas law equation, we can calculate the number of moles of gas in the container:

n = (P * V) / (R * T) = (2.43 atm * 67 L) / (0.0821 L·atm/(mol·K) * 400.15 K) = 0.684 mol

Since there are 3 moles of N2 produced for every 0.289 mol of N2 that we calculated in part (a), we can set up a proportion:

(3 mol N2 / 0.289 mol N2) = (x / 0.684 mol)

Solving for x, x = (3 mol N2 * 0.684 mol) / 0.289 mol N2 = 7.09 mol N2

So, the actual yield of nitrogen gas is 7.09 g. To calculate the percent yield, we can use the formula:

Percent Yield = (Actual Yield / Theoretical Yield) * 100%

Percent Yield = (7.09 g / 8.094 g) * 100% = 87.6%