A sample of hydrogen gas is generated from a closed container (100mL) by reactin

Question

A sample of hydrogen gas is generated from a closed container (100mL) by reacting 3.500g of lithium metal 15.0 mL of 1.00 M nitric acid.
A) write a balanced equation including all phases for this reaction and calculate the number of moles of hydrogen formed.
B) Calculate the partial pressure at 20 degrees Celsius ignoring any solubility of the gas in the solution.
C) the Henry's Law Constant for hydrogen gas in water at 20 degrees Celsius is 3.7 x 10^-4 mol/L x atm. Estimate the number of moles of hydrogen gas that remain dissolved in the solution. What fraction of the gas molecules in the system is dissolved in the solution? Was it reasonable to ignore any dissolved hydrogen in part b? A sample of hydrogen gas is generated from a closed container (100mL) by reacting 3.500g of lithium metal 15.0 mL of 1.00 M nitric acid.
A) write a balanced equation including all phases for this reaction and calculate the number of moles of hydrogen formed.
B) Calculate the partial pressure at 20 degrees Celsius ignoring any solubility of the gas in the solution.
C) the Henry's Law Constant for hydrogen gas in water at 20 degrees Celsius is 3.7 x 10^-4 mol/L x atm. Estimate the number of moles of hydrogen gas that remain dissolved in the solution. What fraction of the gas molecules in the system is dissolved in the solution? Was it reasonable to ignore any dissolved hydrogen in part b?
A) write a balanced equation including all phases for this reaction and calculate the number of moles of hydrogen formed.
B) Calculate the partial pressure at 20 degrees Celsius ignoring any solubility of the gas in the solution.
C) the Henry's Law Constant for hydrogen gas in water at 20 degrees Celsius is 3.7 x 10^-4 mol/L x atm. Estimate the number of moles of hydrogen gas that remain dissolved in the solution. What fraction of the gas molecules in the system is dissolved in the solution? Was it reasonable to ignore any dissolved hydrogen in part b?

Explanation / Answer

A) 2 Li + 2 HNO3 = 2 LiNO3 + H2

   2 x 6.94 gm = 22.4 litre H2 gas

so, 3.500 gm Li = 5.6484 litre H2 gas

Since 22.4 litre H2 gas = 1 mole

so 5.6484 litre H2 gas = 0.252 mole

(B) PH2 = nRT

= 0.252 x 0.0821 x 293 = 6.0619 atm

(C) Henry Law

C=kPgas

= 3.7 x 10-4 x 6.0619 = 22.4 x 10-4

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