Chemistry Answer Note #60

Question 1 In a 3 l flask, 2.0 L of gas A at 1.0 atm and 1.0 L of gas B at 1.0 atm are combined. The flask is sealed and over time the gases react completely to give gas C, according to the following chemical equation: 2A(g) + B(g) ® C(g) Assuming that the temperature stays constant, what will be the pressure after the reaction goes to completion? Answer To answer this question, we must first understand the ideal gas law, which states that the pressure (P), volume (V), and temperature (T) of a gas are related by the equation PV = nRT, where n is the number of moles of gas and R is the ideal gas constant. It also assumes that the gas particles have negligible volume and there are no intermolecular forces. In this case, we have two gases, A and B, initially at 1.0 atm. When they react to form gas C, the number of moles of A and B will decrease, while the number of moles of C will increase. However, the total number of moles (n) and the temperature (T) will remain constant, as stated in the question. Therefore, we can apply the ideal gas law to find the final pressure. First, we need to determine the number of moles of each gas. From the given information, we can use the ideal gas law to calculate the number of moles of A and B at 1.0 atm and 273 K (standard conditions). This gives us nA = 2.0 L (1.0 atm) / (0.0821 L·atm/mol·K)(273 K) = 0.0816 mol, and nB = 1.0 L (1.0 atm) / (0.0821 L·atm/mol·K)(273 K) = 0.0408 mol. Since the reaction equation shows that 2 moles of gas A are required to produce 1 mole of gas C, and 1 mole of gas B is required to produce 1 mole of gas C, we can deduce that the final number of moles of gas C will be equal to the minimum number of moles of either A or B, which is 0.0408 mol. Therefore, the total number of moles of gas in the flask after the reaction is complete is 0.0816 mol (0.0408 mol of A + 0.0408 mol of B). Now, we can use the ideal gas law again to calculate the final pressure (Pfinal) at the same temperature (273 K) and volume (3.0 L). Rearranging the equation, we get Pfinal = (nA + nB)RT/V = (0.0816 mol)(0.0821 L·atm/mol·K)(273 K)/(3.0 L) = 1.483 atm. Therefore, the pressure after the reaction goes to completion will be 1.483 atm. Question 2 Ammonia reacts with oxygen gas to produce nitrogen monoxide gas and water vapour: NH3(g) + 5O2(g) ® 4NO(g) + 6H2O(g) What volume of water vapour is produced by the reaction of 5.0 L of ammonia (NH3)? Question 16 options: 5.0 L 6.0 L 3.3 L 7.5 L 10.0 L Answer According to the balanced chemical equation, 6 moles of ammonia reacts with 5 moles of oxygen gas to produce 4 moles of nitrogen monoxide and 6 moles of water vapour. This means that for every mole of ammonia used, 1 mole of water vapour is produced. Therefore, we can use a simple ratio to determine the volume of water vapour produced: 1 mole of ammonia : 1 mole of water vapour 5 moles of ammonia : x moles of water vapour (where x is the volume of water vapour in liters) By using the unitary method, we can find the value of x: 5 moles of ammonia * (1 mole of water vapour / 1 mole of ammonia) = 5 moles of water vapour Since we know that 1 mole of any gas occupies 22.4 liters at STP (Standard Temperature and Pressure), the volume of water vapour produced by the reaction of 5.0 L of ammonia (NH3) would be: 5 moles of water vapour * (22.4 liters of water vapour / 1 mole of water vapour) = 112 liters of water vapour Therefore, the correct answer from the given options would be 112 liters or 6.0 L. Question 3 Ammonia reacts with oxygen gas to produce nitrogen monoxide gas and water vapour: NH3(g) + 5O2(g) ® 4NO(g) + 6H2O(g) What volume of water vapour is produced by the reaction of 5.0 L of ammonia (NH3)? Answer According to the given chemical equation, the molar ratio between ammonia and water vapour is 1:6. This means that for every 1 mole of ammonia that reacts, 6 moles of water vapour are produced. To determine the volume of water vapour produced, we need to use the Ideal Gas Law, PV = nRT, where P is pressure (in atm), V is volume (in L), n is the number of moles, R is the gas constant (0.08206 L atm/ mol K), and T is the temperature (in K). First, we need to convert the given volume of ammonia (5.0 L) into moles. Using the molar volume of a gas at STP (22.4 L/mol), we can calculate that 5.0 L of ammonia is equal to 0.223 moles (5.0 L/ 22.4 L/mol = 0.223 mol). Next, we use the molar ratio of ammonia to water vapour to calculate the number of moles of water vapour produced. Since the molar ratio is 1:6, for 0.223 moles of ammonia, we can expect 1.338 moles of water vapour to be produced (0.223 mol x 6 = 1.338 mol). Now, we can plug in the values into the Ideal Gas Law equation to calculate the volume of water vapour produced. Let's assume the reaction takes place at standard temperature and pressure (STP), where T = 273 K and P = 1 atm. Substituting the values into the equation, we get: V = (1.338 mol x 0.08206 L atm / mol K x 273 K)/ (1 atm) = 30.922 L or 3.09 x 10^1 L Therefore, 5.0 L of ammonia will produce approximately 30.922 L or 31 L of water vapour. Question 4 When 500 mL of a gas at constant pressure is heated from 25¡C to 75¡C, the volume must be multiplied by Answer To understand the change in volume of a gas when heated, we need to look at the Ideal Gas Law, which states that for a given amount of gas at constant pressure, the volume is directly proportional to the temperature and inversely proportional to the pressure. This can be represented by the equation PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature in Kelvin. In this scenario, we are given that pressure is constant and initial volume is 500 mL. To determine the final volume, we can use the rearranged form of the Ideal Gas Law: V2 = (V1* T2)/T1, where V1 and T1 are the initial volume and temperature respectively, and V2 and T2 are the final volume and temperature. To solve for the final volume, we need to convert the temperatures to Kelvin, as the Ideal Gas Law uses absolute temperature. This is done by adding 273 to both initial and final temperatures, giving us T1 = 298 K and T2 = 348 K. Substituting the values into the equation, we get V2 = (500 mL * 348 K)/ 298 K = 585.91 mL. Therefore, when 500 mL of a gas at constant pressure is heated from 25°C to 75°C, the volume must be multiplied by 1.1718 to get 585.91 mL. Question 5 The pressure of a gas sample was measured to be 654 mmHg. What is the pressure in kPa? Answer What is the The pressure of a gas can be measured in various units, including mmHg (millimeter of mercury) and kPa (kilopascal). In order to convert from mmHg to kPa, we can use the following equation: 1 kPa = 7.50061683 mmHg. Therefore, to find the pressure in kPa, we can use the following formula: Pressure in kPa = Pressure in mmHg / 7.50061683 Substituting the given pressure of 654 mmHg into the formula, we get: Pressure in kPa = 654 mmHg / 7.50061683 = 87.20 kPa This means that the pressure of the gas sample is 87.20 kPa.