Question

In the reaction potassium fluoride with potassium chromate and sulfuric acid, the products are fluorine gas, potassium sulfate, chromium (III) sulfate and water. If 39.3 liters of fluorine gas was formed in this reaction and the gas was collected at -10.0 degrees C and 725 Torr, how many grans of potassium fluoride reacted?

Answer 1

reaction is

6 KF + 2 K₂CrO₄ + 8 H₂SO₄ ...................>  3 F₂ + Cr₂(SO₄)₃ + 8 H₂O + 5 K₂SO₄

6 mole KF produce 3 mole F2 gas.

V = 39.3 L

T = 273 - 10.0 = 263 K

P = 725 / 760 = 0.9539 atm

ideal gas equation is

PV = nRT

0.9539 * 39.3 = n * 0.0821 * 263

or

n = 1.736 mole.

thus

mole of F2 gas produced = 1.736 mole.

and

mole of KF used = (6 / 3) * 1.736 = 3.472 mole

and

grans of potassium fluoride reacted = 3.472 mole * 58.0967 g/mole = 201.7 g (answer)