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How do you calculate the volume in liters of 1.50 mol Cl2 at stp?
Wiki User
∙ 13y ago
Use PV =nRT ( pressure at STP is 1 atmosphere and temp. is 298.15 Kelvin )
(1 atm)(volume) = (1.50 mole)(0.08206 Latm/molK)(298.15 K)
= 36.7 Liters
Wiki User
∙ 13y ago
Wiki User
∙ 9y agoApprox. 1 mol.
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How many liters of chlorine gas can be produced when 0.98L of HCl react with excess O2 at STP?
the ratio between HCL and CL2 is 4:2 V(Cl2)=0.98/2=0.49L
What volume is contained of 2.4 moles of chlorine and how do you work it out?
I would assume chlorine gas and standard temperature an atmospheric pressure. Using the ideal gas equation. PV = nRT (1 atm)(X volume) = (2.4 moles Cl2)(0.08206 Mol*K/L*atm)(298.15 K) Volume = 59 Liters of chlorine gas --------------------------------------------
How many liters are in 4.50 moles of chlorine gas?
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.
For the reaction Cl2 plus 2KBr equals Br2 calculate the percentage yield if 200 g of Cl2 react with excess KBr to produce 410 g of Br2?
91%
A reaction of 1 liter of chlorine gas Cl2 with 3 liters of fluorine gas F2 yields 2 liters of a gaseous product All gas volumes are at the same temperature and pressure What is the formula?
ClF3
How many liters of 17.32g of chlorine gas can be produced?
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 volume occupied by 30lb of Cl2 at 743 mm Hg and 70 degree F?
Some conversion required.30 pounds Cl2 (454 grams/1 pound) = 13620 grams Cl213620 grams Cl2 (1 mole Cl2/70.9 grams) = 192.1 moles Cl270o F = 21.1o C = 294.3 Kelvinuse,PV = nRT(743 mm Hg)(X Volume) = (192.1 moles Cl2)(62.36 L*torr/mol*K)(294.3 K)Volume = 3525524.47/743= 4.7 X 103 Liters of chlorine gas------------------------------------------
Volume of 25 g of Cl2?
0.016L
How do you calculate the balance equation for Na plus Cl2?
Balanced equation for Sodium and Chlorine is: 2Na + Cl2 --> 2NaCl
What volume will the Cl2 occupy at STP?
stupid question
How many liters of chlorine gas can be produced when 0.98L of HCl react with excess O2 at STP?
the ratio between HCL and CL2 is 4:2 V(Cl2)=0.98/2=0.49L
What volume is contained of 2.4 moles of chlorine and how do you work it out?
I would assume chlorine gas and standard temperature an atmospheric pressure. Using the ideal gas equation. PV = nRT (1 atm)(X volume) = (2.4 moles Cl2)(0.08206 Mol*K/L*atm)(298.15 K) Volume = 59 Liters of chlorine gas --------------------------------------------
How many liters are in 4.50 moles of chlorine gas?
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.
For the reaction Cl2 plus 2KBr equals Br2 calculate the percentage yield if 200 g of Cl2 react with excess KBr to produce 410 g of Br2?
91%
A reaction of 1 liter of chlorine gas with 3 liters of Fluorine gas yields 2 liters of a gaseous produce What is the formula of the gaseous product?
Cl2)+ 3 F2 = 2 ClF3
A reaction of 1 liter of chlorine gas Cl2 with 3 liters of fluorine gas F2 yields 2 liters of a gaseous product All gas volumes are at the same temperature and pressure What is the formula?
ClF3
What reactant is limiting if 3000 cm3 of Cl2 at STP react with a solution containing 25 grams of NaBr?
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
