Answer:
Explanation:
Hello.
In this case, for the reaction between aqueous solutions of ammonium chloride and iron (III) hydroxide, we have the following complete molecular reaction:
And the full ionic equation, taking into account that the iron (III) hydroxide cannot be dissolved as it is insoluble in water:
Finally, the net ionic equation, considering that spectator ions are NH₄⁺, Cl⁻ as they are both the left and right side, therefore, the net ionic equation is:
Best regards.
16 g. The mass of 0.60 mol Al is 16 g.
Molar mass of Al = 26.98 g/mol
Mass of Al = 0.60 mol Al x (26.98 g Al/1 mol Al) = 16 g Al
Hey there : !
density = 10.5 g/cm³
volume = 23.6 cm³
therefore:
D = m / V
10.5 = m / 23.6
m = 10.5 * 23.6
m = 247.8 g
hope this helps!
Answer:
0.098 moles H₂S
Explanation:
The reaction that takes place is
- 2H₂(g) + S₂(g) ⇄ 2H₂S(g) keq = 7.5
We can express the equilibrium constant as:
- keq = [H₂S]² / [S₂] [H₂]² = 7.5
With the volume we can <u>calculate the equilibrium concentration of H₂</u>:
- [H₂] = 0.072 mol / 2.0 L = 0.036 M
<em>The stoichiometric ratio</em> tells us that <u>the concentration of S₂ is half of the concentration of H₂</u>:
- [S₂] = [H₂] / 2 = 0.036 M / 2 = 0.018 M
Now we <u>can calculate [H₂S]</u>:
- 7.5 = [H₂S]² / (0.018*0.036²)
So 0.013 M is the concentration of H₂S <em>at equilibrium</em>.
- This would amount to (0.013 M * 2.0 L) 0.026 moles of H₂S
- The moles of H₂ at equilibrium are equal to the moles of H₂S that reacted.
Initial moles of H₂S - Moles of H₂S that reacted into H₂ = Moles of H₂S at equilibrium
Initial moles of H₂S - 0.072 mol = 0.026 mol
Initial moles of H₂S = 0.098 moles H₂S
