How many grams of XeF6 are required to react with 0.579 L of hydrogen gas at 6.46 atm and 45°C in the reaction shown below?
1 answer:
The chemical reaction equation for this is
XeF6 + 3H2 ---> Xe + 6HF
Assuming gas behaves ideally, we use the ideal gas formula to solve for number of moles H2 with T = 318.15K (45C), P = 6.46 atm, V = 0.579L. Then we use the gas constant R = 0.08206 L atm K-1 mol-1.
we get n = 0.1433 moles H2
to get the mass of XeF6,
we divide 0.1433 moles H2 by 3 since 1 mole XeF6 needs 3 moles H2 to react then multiply by the molecular weight of XeF6 which is 245.28 g/mole XeF6.
0.1433 moles H2 x
x 
= 11.71 g XeF6
Therefore, 11.71 g of XeF6 is needed to completely react with 0.579 L of Hydrogen gas at 45 degrees Celcius and 6.46 atm.
XeF6 + 3H2 ---> Xe + 6HF
Assuming gas behaves ideally, we use the ideal gas formula to solve for number of moles H2 with T = 318.15K (45C), P = 6.46 atm, V = 0.579L. Then we use the gas constant R = 0.08206 L atm K-1 mol-1.
we get n = 0.1433 moles H2
to get the mass of XeF6,
we divide 0.1433 moles H2 by 3 since 1 mole XeF6 needs 3 moles H2 to react then multiply by the molecular weight of XeF6 which is 245.28 g/mole XeF6.
0.1433 moles H2 x
Therefore, 11.71 g of XeF6 is needed to completely react with 0.579 L of Hydrogen gas at 45 degrees Celcius and 6.46 atm.
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Answer:
20.3 kJ of heat is absorbed when 9.00 g of steam condenses to liquid water.
Explanation:
Heat is being consumed during vaporization and heat is being released during condensation.
To vaporize 1 mol of water, 40.66 kJ of heat is being consumed.
Molar mass of water = 18.02 g/mol
Hence, to vaporize 18.02 g of water , 40.66 kJ of heat is being consumed.
So, to vaporize 9.00 g of water, of heat or 20.3 kJ of heat is being consumed
As condensation is a reverse process of vaporization therefore 20.3 kJ of heat is absorbed when 9.00 g of steam condenses to liquid water.