Calculate the Volume of Ammonia Gas at Standard Temperature and Pressure
What volume will 3.50 mol of ammonia gas occupy at conditions of standard temperature and pressure?
O A. 5.41 L
OB. 10.5 L
O C. 78.4 L
O D. 7,940 L
What volume will 3.50 mol of ammonia gas occupy at conditions of standard temperature and pressure? Final answer: At standard temperature and pressure, 3.50 mol of ammonia gas will occupy approximately 783.92 L. Explanation: The volume of a gas can be calculated using the ideal gas law, which states that: PV = nRT Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin. At standard temperature and pressure (STP), the pressure is 1 atmosphere (atm) and the temperature is 273 Kelvin (K). From the given information, we know that there are 3.50 mol of ammonia gas. We can substitute the values into the ideal gas law equation: (1 atm) * V = (3.50 mol) * (0.0821 L·atm/mol·K) * (273 K) Simplifying the equation: V = (3.50 mol) * (0.0821 L·atm/mol·K) * (273 K) / (1 atm) V = 783.9225 L Thus, 3.50 mol of ammonia gas will occupy approximately 783.92 L at standard temperature and pressure.