Raoul Pictet the Swiss physicist who first liquefied oxygen, attempted to liquef

Question

Raoul Pictet the Swiss physicist who first liquefied oxygen, attempted to liquefy hydrogen. He heated potassium formate, KCHO2 with KOH in a closed 2.50 Liter vessel.
KCHO2(s) + KOH(s) = K2CO3(s)+H2(g)
If 75.0g of potassium formate reacts in a 2.50 Liter vessel which was initially evacuated, what pressure of hydrogen will be attained when the temperature is finally cooled to 25 degrees Celsius? Use the preceding chemical equation and ignore the volume of solid product Raoul Pictet the Swiss physicist who first liquefied oxygen, attempted to liquefy hydrogen. He heated potassium formate, KCHO2 with KOH in a closed 2.50 Liter vessel.
KCHO2(s) + KOH(s) = K2CO3(s)+H2(g)
If 75.0g of potassium formate reacts in a 2.50 Liter vessel which was initially evacuated, what pressure of hydrogen will be attained when the temperature is finally cooled to 25 degrees Celsius? Use the preceding chemical equation and ignore the volume of solid product
KCHO2(s) + KOH(s) = K2CO3(s)+H2(g)
If 75.0g of potassium formate reacts in a 2.50 Liter vessel which was initially evacuated, what pressure of hydrogen will be attained when the temperature is finally cooled to 25 degrees Celsius? Use the preceding chemical equation and ignore the volume of solid product

Explanation / Answer

initial KCHO2 = 75 g/84.11 g/mol = 0.892 moles

1 mole of KCHO2 gives 1 mole of H2 gas

So,

moles of H2 formed = 0.892 moles

Temperature (T) = 25 oC + 273 = 298 K

Volume (V) = 2.50 L

R = gas constant

Pressure of H2 gas (P) = ?

Using,

P = nRT/V

   = 0.892 x 0.08205 x 298/2.50

   = 8.72 atm

Pressure of hydrogen gas attained is 8.72 atm

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