Answer:
648.68 mg
Explanation:
The reaction that takes place is:
- FeCl₃ + 3NaOH → Fe(OH)₃ + 3NaCl
First we<u> calculate how many moles of each reactant were added</u>, using the <em>given volumes and concentrations</em>:
- FeCl₃ ⇒ 100 mL * 0.240 M = 24 mmol FeCl₃
- NaOH ⇒ 100 mL * 0.182 M = 18.2 mmol NaOH
24 mmol of FeCl₃ would react completely with (24 * 3) 72 mmol of NaOH. There are not as many NaOH mmoles, so NaOH is the limiting reactant.
Now we <u>calculate how many moles of Fe(OH)₃ are formed</u>, using the <em>moles of the limiting reactant</em>:
- 18.2 mmol NaOH *
= 6.07 mmol Fe(OH)₃
Finally we <u>convert 6.07 mmol Fe(OH)₃ to grams</u>, using its<em> molar mass</em>:
- 6.07 mmol Fe(OH)₃ * 106.867 mg/mmol = 648.68 mg
<u>Answer:</u> The net ionic equation is written below.
<u>Explanation:</u>
Net ionic equation of any reaction does not include any spectator ions.
Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.
The chemical equation for the reaction of magnesium nitrate and aqueous ammonia (ammonium hydroxide) is given as:
A white precipitate of magnesium hydroxide is formed in the above reaction.
Ionic form of the above equation follows:
As, ammonium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.
The net ionic equation for the above reaction follows:
Hence, the net ionic equation is written above.
