I am probably wrong but I would think 900. I wouldn’t submit that without somebody backing me up though.
Answer: the answer is D im pretty sure
Explanation:
3 moles FeCl2
Explanation:
For any chemical reaction, the balanced chemical equation tells you the ratio that must always exist between the reactants.
In your case, you have
FeCl2(aq) + 2NaOH(aq) → Fe(OH)2(s) ↓ + 2 NaCl(aq)
The stoichiometric coefficients that belong to iron(II) chloride and to sodium hydroxide, respectively, tell you the mole ratio that must exist between the two reactants when this reaction takes place.
Notice that you have a 1:2 mole ratio between the two reactants, so you can say that the reaction will always consume twice as many moles of sodium hydroxide than moles of iron(II) chloride.
Now, you know that 6 moles of iron(II) chloride are added to 6 moles of sodium hydroxide.
Use the aforementioned mole ratio to determine how many moles of iron(II) chloride will react with the moles of sodium hydroxide
6 moles NaOH ⋅ 1 mole FeCl2
———————
2 moles NaOH
= 3 moles FeCl 2
This tells you that in order for all the moles of sodium hydroxide to react, you need 3 moles of iron(II) chloride. The other 3 moles will not take part in the reaction, i.e. they are in excess.
So, you can say that
3 moles of FeCl 2 → will react
3 moles of FeCl 2 → will not react
Notice that sodium hydroxide is completely consumed before all the moles of iron(II) chloride get the chance to take part in the reaction.
This tells you that sodium hydroxide acts as a limiting reagent, i.e. it limits the amount of iron(II) chloride that takes part in the reaction from 6 moles to 3 moles.
Answer is: Ka for the
monoprotic acid is 3.03·10⁻⁴.
<span>
Chemical reaction: HA(aq) </span>⇄ A⁻(aq) + H⁺<span>(aq).
c(monoprotic acid) = 0.0165 M.
pH = 2.68.
[A</span>⁻] = [H⁺] = 10∧(-2.68).<span>
[A</span>⁻]
= [H⁺] = 0.00209 M; equilibrium concentration.<span>
[HA] = 0.0165 M - 0.00209 M.
[HA] = 0.0144 M.
Ka = [A</span>⁻]·[H⁺] / [HA].<span>
Ka = (0.00209 M)² / 0.0144 M.
Ka = 0.000303 M = 3.03·10</span>⁻⁴<span> M.</span>
