17.7
Cl2 + 2NaBr => Br2 + 2NaCl One mole Cl2 reacts with 2 moles NaBr Cl2 = 71 NaBr = 102.9 Molar volume = 22.414 L/mole for ideal gas @STP 3L Cl2 = 3/22.414 = 0.1338 mole 25g NaBr = 25/102.9 = 0.2430 mole 0.1338 moles Cl2 requires 0.2676 moles NaBr for complete reaction The NaBr is the limiting reagent
Approx 0.223 moles.
3.058
Since you don't specify the conditions, I will use the current STP. At STP of 0oC (273.15 K for gases) and 100 kPa, the molar volume of a gas is 22.711 L/mol.4.50 mol Cl2 x 22.711 L/mol = 102 L Cl2, rounded to three significant figures.Just in case your teacher is using the older version of STP of 0oC (273.15 K for gases), and 1 atm, the molar volume of a gas would be 22.414 L/mol.4.50 mol Cl2 x 22.414 L/mol = 101 L Cl2, rounded to three significant figures.
4,54 L of CO2 have 0,204 moles.
0.250 mol
Cl2 + 2NaBr => Br2 + 2NaCl One mole Cl2 reacts with 2 moles NaBr Cl2 = 71 NaBr = 102.9 Molar volume = 22.414 L/mole for ideal gas @STP 3L Cl2 = 3/22.414 = 0.1338 mole 25g NaBr = 25/102.9 = 0.2430 mole 0.1338 moles Cl2 requires 0.2676 moles NaBr for complete reaction The NaBr is the limiting reagent
At STP, Cl2 gas has a density of 0.0032g/mL. (Wikipedia article, "Chlorine") 0.30g Cl2 x (1mL Cl2/0.0032g) = 93.75mL Cl2
At 22.4 liters a mole at STP, and the molar mass of Cl2 being 71, 17.32 g is about .242 moles. Multiply the moles by standard volume and you get 5.43 liters.
what is the volume of a balloon containing 50.0 moles of O2 gas at a pressure of 15.0 atm at 28 degrees
0.25 moles
8,4 liters of nitrous oxide at STP contain 2,65 moles.
16,8 L of Xe gas at STP is equivalent to 0,754 moles.
The answer is 0,2675 moles.
The answer is 2,68 moles.
Approx 0.223 moles.
stupid question