Answer:
- 5.15×10²⁴ molecules of sulfur dioxide
- 3.63×10²³ molecules of carbon monoxide
- 6.02×10²³ molecules of ammonia
Explanation:
We begin from the relation that 1 mol of molecules contains NA of molecules
NA = 6.02×10²³
Now, we make rules of three:
1 mol has 6.02×10²³ molecules, therefore:
8.55 moles of SO₂ must have (8.55 . NA) / 1 = 5.15×10²⁴ molecules of dioxide
0.603 moles of CO must have (0.603 . NA) / 1 = 3.63×10²³ molecules of monoxide
Avogadro's Number of molecules of NH₃ are 6.02×10²³ molecules of ammonia
Answer:
option c is correct
Explanation:
the addition of catalyst does not effect the position or equilibrium constant and increase the forward and backward reaction in equal rates so no effect would be observed
We have Kc = 4.2 x 10^-2 (given but missing in the question)
and When the balanced equation for this reaction is:
PCl5(g) ↔ PCl3(g) + Cl2(g)
so, according to the Kc formula:
Kc = the concentration of products / the concentration of the reactants
so, to get the concentration of the reactants in equilibrium, the concentration of the products / the concentration of the reactants should equal the Kc value which is given in the question (missing in your question).
So by substitution in Kc formula:
Kc = [PCl3]*[Cl2] / [PCl5]
4.2 x 10^-2 = 0.18 * 0.25 /[PCl5]
∴[PCl5] = 0.18*0.25 / 4.2x10^-2 = 1.07
So the concentration of the reactants in equilibrim = 1.07
