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rusak2 [61]
3 years ago
7

Write the net ionic equation for the reaction of aqueous solutions of ammonium chloride and iron(III) hydroxide.

Chemistry
1 answer:
vagabundo [1.1K]3 years ago
8 0

Answer:

Fe(OH)_3(s)\rightarrow Fe^{3+}(aq)+3OH^-(aq)

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:

3NH_4Cl(aq)+Fe(OH)_3(s)\rightarrow 3NH_4OH+FeCl_3

And the full ionic equation, taking into account that the iron (III) hydroxide cannot be dissolved as it is insoluble in water:

3NH_4^+(aq)+3Cl^-(aq)+Fe(OH)_3(s)\rightarrow 3NH_4^+(aq)+3OH^-(aq)+Fe^{3+}(aq)+3Cl^-(aq)

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:

Fe(OH)_3(s)\rightarrow Fe^{3+}(aq)+3OH^-(aq)

Best regards.

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N2+3H2 → 2NH3
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Explanation:

N2 (g) + H2 (g) gives out NH3 (g)

Now balance it. You have two reactants with compositions involving a single element, which makes it very easy to keep track of how much is on each side. I would balance the nitrogens, and then the hydrogens.

Now balance it. You have two reactants with compositions involving a single element, which makes it very easy to keep track of how much is on each side. I would balance the nitrogens, and then the hydrogens.(If you balance the hydrogen reactant with a whole number first, I can guarantee you that you will have to give NH3 a new stoichiometric coefficient.)

N2 (g) + 3H2 (g) gives out 2NH3 (g)

The stoichiometric coefficients tell you that if we can somehow treat every component in the reaction as the same (like on a per-mol basis, hinthint), then one "[molar] equivalent" of nitrogen yields two [molar] equivalents of ammonia.

Luckily, one mol of anything is equal in quantity to one mol of anything else because the comparison is made in the units of mols.

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N2.

After doing the actual calculation you should realize that we have about 4 times as much H2 as we need. Therefore the limiting reagent is clearly N2.

Thus, we should yield 2×0.9995=1.9990 mols of NH3 (refer back to the reaction). So this is the second and last calculation we need to do:

1.9990 mol NH3 × 17.0307 g NH3/ 1 mol NH3

= 34.0444 g NH3

Hope it helpz~

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