Asked by caroline
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 L vessel.
KCHO2(s) + KOH(s) → K2CO3(s) + H2(g)
If 50.1 g of potassium formate reacts in a 2.50-L vessel, which was initially evacuated, what pressure of hydrogen will be attained when the temperature is finally cooled to 25°C? Use the preceding chemical equation and ignore the volume of solid product.
KCHO2(s) + KOH(s) → K2CO3(s) + H2(g)
If 50.1 g of potassium formate reacts in a 2.50-L vessel, which was initially evacuated, what pressure of hydrogen will be attained when the temperature is finally cooled to 25°C? Use the preceding chemical equation and ignore the volume of solid product.
Answers
Answered by
caroline
what equation would you have to use here?
Answered by
DrBob222
I would try converting KCHO2 to moles. 50.1 g/molar mass = moles, then convert that to M = moles/L (divide moles by 2.5L).
Since 1 mole H2 is formed from 1 mole KCHO2, then use PV = nRT to solve for P. All of the other reactants and products are solids and the problem tells you to ignore the solids.
Since 1 mole H2 is formed from 1 mole KCHO2, then use PV = nRT to solve for P. All of the other reactants and products are solids and the problem tells you to ignore the solids.
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