I think it would be Bromine and Mercury, hope that helps
The given reaction is a combustion reaction, since a hydrocarbon is burning in presence of oxygen
<h3><u>Answer</u>;</h3>
= 226 Liters of oxygen
<h3><u>Explanation</u>;</h3>
We use the equation;
LiClO4 (s) → 2O2 (g) + LiCl, to get the moles of oxygen;
Moles of LiClO4;
(500 g LiClO4) / (106.3916 g LiClO4/mol)
= 4.6996 moles
Moles of oxygen;
But, for every 1 mol LiClO4, two moles of O2 are produced;
= 9.3992 moles of Oxygen
V = nRT / P
= (9.3992 mol) x (8.3144621 L kPa/K mol) x (21 + 273) K / (101.5 kPa)
= 226 L of oxygen
The reaction is:
2 KClO3(s) → 3 O2(g) + 2 KCl(s) <span>
<span>A catalyst simply lowers the activation energy so MnO2 is not
part of the overall reaction.
By stoichiometry:
<span>3.45 g KClO3 x 1 mol/ 122.55g KClO3 x 3 mol O2/ 2 mol KClO3 x
31.99g/ 1 mol O2 = 331.096/ 245.1 = 1.35 grams O2 produced
Answer:1.35 grams O2</span></span></span>
