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### Quimica

Posted: Tue Nov 09, 2021 1:40 pm
Considere la siguiente ecuación química:
2KMnO4(ac) + 10KCl(ac) + 8H2SO4(l) → 2MnSO4(ac) + 6K2SO4(ac) + 8H2O(l) + 5Cl2(g)

Al hacer reaccionar 120 [g] de permanganato de potasio en presencia de 50 [g] de cloruro de potasio en medio acuoso ácido (ácido sulfúrico en exceso), únicamente se colectó el 65% de cloro.

Determine el volumen, en litros, de cloro producido si la reacción se llevó a cabo a 25 [ºC] y 1 [atm].

### Re: Quimica

Posted: Tue Nov 09, 2021 2:24 pm
Translation in English
Consider the following chemical equation:
2KMnO4 (aq) + 10KCl (aq) + 8H2SO4 (l) → 2MnSO4 (aq) + 6K2SO4 (aq) + 8H2O (l) + 5Cl2 (g)

By reacting 120 [g] of potassium permanganate in the presence of 50 [g] of potassium chloride in acidic aqueous medium (excess sulfuric acid), only 65% ​​of chlorine was collected.

Determine the volume, in liters, of chlorine produced if the reaction was carried out at 25 [° C] and 1 [atm].

### Re: Quimica

Posted: Tue Nov 09, 2021 2:39 pm
You have 1 mol Permanganate to 5 mol Chloride to 2,5 mol Chlorine
Convert both given mass to mol and compare what is in excess and dearth. The lower value drives the reaction. From this you calculate mol of chlorine.
65% of this value is the mole what has to be converted to the volume. Use ideal Gas equation.