Final answer:
The volume of 35.2g of methane gas at 1.00 atm and 25 °C, convert the mass to moles, use the Ideal Gas Law with the universal gas constant, and then solve for volume. The calculation shows that 35.2g of methane occupies 53.3 liters under the given conditions.
Step-by-step explanation:
To calculate the volume occupied by 35.2g of methane gas at 1.00 atm and 25 °C, we need to use the Ideal Gas Law, which is PV=nRT. Here, P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant, and T is the temperature in Kelvin. First, we must find the number of moles of methane by using the formula n = mass / molar mass. For methane (CH₄), the molar mass is 16.0 g/mol, so the number of moles of methane is 35.2g / 16.0 g/mol = 2.2 mol. Using the Ideal Gas Law, we convert the temperature to Kelvin by adding 273.15 to the Celsius temperature, T = 25 °C + 273.15 = 298.15 K.
The universal gas constant, R, is 0.0821 L·atm/K·mol. Now we can rearrange the Ideal Gas Law to solve for V, which gives us V = nRT/P. Plugging in the values, we get V = (2.2 mol)(0.0821 L·atm/K·mol)(298.15 K) / (1.00 atm), which calculates to approximately 53.3 liters. Therefore, 35.2g of methane gas occupies 53.3 liters at 1.00 atm and 25 °C.