These are dissolved in water to form colourless solutions, and then mixed together. This mixing leads to a double displacement reaction, essentially resulting in the metals 'swapping' their places in the two compounds, producing lead (II) iodide, and potassium nitrate.
Answer:
2 KClO3 (s) = 2 KCl (s) + 3 O2 (g)
2.5 g x g
Explanation:
x g O2 = 2.5 g KClO3 x (1 mol KClO3) x (3 mol O2) x (32 g O2) = 0.98 g O2
(122.5 g KClO3) (2 mol KClO3) (1 mol O2)
2 KClO3 (s) 2 KCl (s) + 3 O2 (g)
2.5 g x g
x g KCl = 2.5 g KClO3 x (1 mol KClO3) x (2 mol KClO3) x (74.5 g KCl) = 1.52 g KCl
(122.5 g KClO3) (2 mol KClO3) (1 mol KCl)
2 KClO3 (s) 2 KCl (s) + 3 O2 (g)
x mol 10 mol
x mol KClO3 = 10 mol O2 x (2 mol KClO3) = 6.7 mol KClO3
(3 mol O2)
Answer:
Explanation:
First, we find in the tables the ΔH of formation of each compound. As you can see in the (image 1)
Then we solve the ecuation for ΔH°reaction
ΔH°reaction=∑ΔH°f(products)−∑ΔH°f(Reactants)
ΔH°reaction= (-2* 393.5 - 2*285.8) - (52.4 + 0) kJ/mol
ΔH°reaction = -1.41 *10^3 kJ/mol
The rate of a reaction with this rate law would increase by a factor of 4 if NO concentration were doubled.
