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>
Answer:
i don’t understand what you are saying?
Explanation:
THe addition of strong electrolyte such as MgSO4 compound will led the equilibrium to shift due to common ion effect. The addition of common ion decrease the solubility as the reaction shift to the left thus less product is formed.
<h3>Answer:</h3>
Mole = 0.055 mol
<h3>Solution:</h3>
Step 1: Calculate Mass of Glucose in 200 mL solution:
As,
100 mL Solution contains = 5 g of Glucose
So,
200 mL Solution will contain = X g of Glucose
Solving for X,
X = (200 mL × 5 g) ÷ 100 mL
X = 10 g of Glucose
Step 2: Calculate Moles for 10 g of Glucose:
As,
Mole = Mass ÷ M.Mass
M.Mass of Glucose = 180 g.mol⁻¹
So,
Mole = 10 g ÷ 180 g.mol⁻¹
Mole = 0.055 mol
