volume is 3.85 cubic dm. or litre.....
At STP conditions (standard temperature and pressure), the volume occupied by 1 mole of ideal gas is 22.4 liters. Since the molar mass of SO2 is approximately 64 g/mol, 11 g of SO2 is about 0.172 moles. Therefore, the volume of 11 g of SO2 at STP would be approximately 3.85 liters.
1 mole of gas occupies 22.4 liters at STP. Therefore 3.5/22.4 = 0.15625 moles of SO2. There are thus 0.15625 moles of O2 needed to react with solid sulfur because S + O2 ---->SO2. 0.15625 moles of oxygen occupies 0.15625 x 22.4 liters = 3.5 liters O2 required.
To find the volume of a gas such as sulfur dioxide with a given mass, you need to know the temperature, pressure, and molar mass. Assuming standard temperature and pressure conditions (STP), the volume of 72.0 grams of sulfur dioxide can be calculated using the ideal gas law 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.
The ideal gas law states that at STP (Standard Temperature and Pressure), one mole of gas occupies 22.4 L of volume. Boron trifluoride (BF3) has a molar mass of 67.8 g/mol. Therefore, 0.155 g of boron trifluoride is 0.155/67.8 = 0.00228 moles. At STP, this amount of boron trifluoride would occupy 0.00228 * 22.4 = 0.051072 L of volume.
The volume of hydrogen is 97, 86 L.
At STP conditions (standard temperature and pressure), the volume occupied by 1 mole of ideal gas is 22.4 liters. Since the molar mass of SO2 is approximately 64 g/mol, 11 g of SO2 is about 0.172 moles. Therefore, the volume of 11 g of SO2 at STP would be approximately 3.85 liters.
1 mole of gas occupies 22.4 liters at STP. Therefore 3.5/22.4 = 0.15625 moles of SO2. There are thus 0.15625 moles of O2 needed to react with solid sulfur because S + O2 ---->SO2. 0.15625 moles of oxygen occupies 0.15625 x 22.4 liters = 3.5 liters O2 required.
1 mole SO2 weights 32+2*16 = 64 gram S + 2*O and it's volume is 22.4 l You can do with that to calculate density (g/L) ?
Using the ideal gas law, we can calculate that 0.640 g of SO2 at STP is approximately 0.0286 moles. The balanced chemical equation for the reaction 2SO2 + O2 → 2SO3 indicates that 1 mole of SO2 reacts with 1 mole of O2. Therefore, 0.0286 moles of SO2 would require 0.0286 moles of O2. At STP, 1 mole of gas occupies 22.4 L, which is equivalent to 22,400 mL. So, 0.0286 moles of O2 would occupy 643 mL.
At STP (standard temperature and pressure), 1 mole of any gas occupies 22.4 liters. The molar mass of SO2 is 64 grams/mol. Thus, for 576 grams of SO2, you first need to find the number of moles (576 g / 64 g/mol) and then use the ideal gas law to find the volume: V = (nRT) / P, where n is the number of moles, R is the ideal gas constant, T is the temperature, and P is the pressure.
The balanced equation for the reaction is: 2SO2 + O2 -> 2SO3. Therefore, 1 mole of O2 is needed to react with 2 moles of SO2 to form 2 moles of SO3. So for 200 moles of SO2, you would need 100 moles of O2. At STP, 1 mole of any gas occupies 22.4 L, so the volume of O2 needed would be 2240 L (100 moles x 22.4 L).
The volume of 10.9 mol of helium at STP is 50 litres.
At standard temperature and pressure (STP), the gas that occupies the highest volume is hydrogen.
See the Related Question "How do you solve Ideal Gas Law problems?" to the left for the answer.
The molar volume of hydrogen gas at STP (Standard Temperature and Pressure) is 22.4 liters per mole.
At STP (standard temperature and pressure), one mole of any gas occupies a volume of 22.4 liters. This is known as the molar volume of a gas at STP.
To calculate the volume of CO2 at STP (Standard Temperature and Pressure), you can use the ideal gas law equation: PV = nRT. First, find the number of moles of CO2 using the ideal gas law equation. Then, use the molar volume of a gas at STP (22.4 L/mol) to find the volume at STP.