STP (standard temperature and pressure)
The reaction : Mg + O2-----------> MgO The balanced reaction : 2 Mg + O2-----------> 2 MgO that is 2 moles of Mg reacts with one mole of oxygen gas. moles of Mg reacted = 12 g / 24 g mol-1=0.5 mol. Thus the no. of mol of O2 reacted = 0.5 mol /2 =0.25 mol If the gas is at standard temperature and pressure, then Volume of 1 mol of O2 gas = 22.4 dm3 Volume of 0.25 mol of O2 gas = 22.4 x 0.25 dm3= 5.6 dm3 the volume of oxygen that will be used to react completely with 12 g Mg =5.6 dm3 (provided that the gas is at the standard conditions.)
At STP, 1 mol or 6.02x10^23 representative particles, of any gas occupies a volume of 22.4 Liters. (chemistry)
The molar volume at STP(22.4 L/mol) can be used to calculate the molar mass of the gas.
9g Al x (1 mol Al/27g Al) x (3mol H2 / 2 mol Al) x (22.4L / 1 mol H2) = 11.2L H2
Acetylene is C2H2, with a molar mass of 26g/mol. 49.6g of it = 1.9 moles. At STP, 1 mole of any gas occupies a volume of 22.4 liters, so 1.9 moles at STP would have a volume of 42.56 liters.
1 mole gas = 22.4L 1.5mol C2H4 x 22.4L/mol = 33.6L ethane gas (C2H4)
1 mol of any gas has a volume of 22.4 L at STP
The reaction : Mg + O2-----------> MgO The balanced reaction : 2 Mg + O2-----------> 2 MgO that is 2 moles of Mg reacts with one mole of oxygen gas. moles of Mg reacted = 12 g / 24 g mol-1=0.5 mol. Thus the no. of mol of O2 reacted = 0.5 mol /2 =0.25 mol If the gas is at standard temperature and pressure, then Volume of 1 mol of O2 gas = 22.4 dm3 Volume of 0.25 mol of O2 gas = 22.4 x 0.25 dm3= 5.6 dm3 the volume of oxygen that will be used to react completely with 12 g Mg =5.6 dm3 (provided that the gas is at the standard conditions.)
Because oxygen gas (O2) has a molar mass of 32g/mol, 11.3 g * 1/32 mol/g gives about .35 moles. An ideal gas has a volume of 22.4 L/mol at STP, so 11.3 g O2 would have a volume of 7.91 L at STP.
If you are talking about ideal conditions then you can use the ideal gas law: (Pressure)(Volume)=(Number of particles in moles)(R=8.314472 J·K−1·mol−1)(Temperature) or PV=nRT
At STP, 1 mole of an ideal gas occupies 22.4 liters. 0.335 mol x 22.4 L = 7.50 L .................. 1 mol
That's going to depend on the substance, which the question neglects to identify. --------------------------------------------------- The volume of any gas at STP (pressure of 1 ATM & temp.: 0oC) is approximately 22.41 L/mol or 22,410 mL/mol. So you need to find out how much gas you have to begin with (# of moles) to find the volume of the gas at STP.
The equation you will need is: Mol of substance 1 * volume of substance 1 = Mol of substance 2 * volume of substance 2
At STP, 1 mol or 6.02x10^23 representative particles, of any gas occupies a volume of 22.4 Liters. (chemistry)
At Standard Temperature and Pressure (STP), which is defined as 0 degrees Celsius (273.15 Kelvin) and 1 atmosphere pressure, the molar volume of an ideal gas is approximately 22.4 liters/mol. The molar mass of nitrogen gas (Nā) is approximately 28.02 grams/mol. To calculate the density (D) of nitrogen gas at STP, you can use the ideal gas law: ļæ½ = Molar mass Molar volume at STP D= Molar volume at STP Molar mass ā ļæ½ = 28.02 ā g/mol 22.4 ā L/mol D= 22.4L/mol 28.02g/mol ā ļæ½ ā 1.25 ā g/L Dā1.25g/L Therefore, the density of nitrogen gas at STP is approximately 1.25 grams per liter.
The molar volume at STP(22.4 L/mol) can be used to calculate the molar mass of the gas.
The pressure would double in size.