To determine the volume occupied by 61.0 grams of ammonia gas (NH3), we can use the ideal gas law equation: PV = nRT. In this equation, P represents pressure, V is volume, n is the number of moles of the gas, R is the ideal gas constant, and T is temperature in Kelvin.
1. First, we need to convert the mass of ammonia gas (61.0 grams) into moles. To do this, we use the molar mass of ammonia (NH3), which is 17.031 grams/mol. We divide the given mass by the molar mass to get the number of moles:
61.0 grams / 17.031 grams/mol = 3.58 moles
2. Since the question doesn't provide information about pressure or temperature, we can assume standard temperature and pressure conditions (STP) where the pressure is 1 atm and the temperature is 273 K.
3. Substituting the values into the ideal gas law equation and solving for V, we get:
V = (nRT) / P
V = (3.58 moles * 0.0821 L.atm/mol.K * 273 K) / 1 atm
V = 83.29 L
Therefore, the volume occupied by 61.0 grams of ammonia gas at STP would be approximately 83.29 liters.
Among the provided options, the closest one is 80.2 L, which is a reasonable approximation considering the assumptions made regarding pressure and temperature.
