Question

Chlorine gas reacts with fluorine gas to form chlorine trifluoride.
Cl2(g)+3F2(g)→2ClF3(g)
A 1.90 L reaction vessel, initially at 298 K, contains chlorine gas at a partial pressure of 337 mmHg and fluorine gas at a partial pressure of 877 mmHg .

Answer 1

we know that

for gases

PV = nRT

so

Cl2 gas :

(337 / 760) x 1.9 = n x 0.0821 x 298

n = 0.034436

so

moles of chlorine = 0.034436

now

F2 gas :

( 877 / 760) x 1.9 = n x 0.0821 x 298

n = 0.0896

so

moles of F2 = 0.0896

now

consider the given reaction

Cl2 + 3F2 ---> 2ClF3

we can see that

moles of F2 required = 3 x moles of Cl2

so

moles of F2 required = 3 x 0.034436 = 0.103308

but

only 0.0896 moles of F2 is present

so

F2 is the limiting reagent and Cl2 is in excess

now

Cl2 + 3F2 ---> 2 ClF3

we can see that

moles of ClF3 produced = (2/3) x moles of F2

moles of ClF3 produced = 2 x 0.0896 / 3 = 0.05974

now

mass = moles x molar mass

so

mass of ClF3 = 0.05974 x 92.45

mass of ClF3 = 5.523

so

5.523 grams of ClF3 is the theoretical yield