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Schach [20]
3 years ago
10

What is the overall equation for FeCl3(aq) + NH3(aq) + H2O(l)? the overall ionic equation? then the net ionic equation?

Chemistry
2 answers:
diamong [38]3 years ago
4 0
So the question ask to calculate the over all equation for a certain element and the following are 
- over all equation -  FE(OH)3+3NH4CI
- over all ionic equation -  e(OH)3 (s) & 3 NH4+ (aq) & 3Cl- (aq) - net ionic equation -  Fe+3+3OH-+3NH$++3CI-
Amanda [17]3 years ago
4 0

<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 iron(III) chloride, ammonia and water is given as:

FeCl_3(aq.)+3NH_3(g)+3H_2O(l)\rightarrow Fe(OH)_3(s)+3NH_4Cl(aq.)

Ionic form of the above equation follows:

Fe^{3+}(aq.)+3Cl-(aq.)+3NH_3(g)+3H_2O(l)\rightarrow Fe(OH)_3(s)+3NH_4^+(aq.)+3Cl^-(aq.)

As, chloride ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and is the spectator ion.

The net ionic equation for the above reaction follows:

Fe^{3+}(aq.)+3NH_3(g)+3H_2O(l)\rightarrow Fe(OH)_3(s)+3NH_4^+(aq.)

Hence, the net ionic equation is written above

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How many moles of MgO are produced when .250 mol of Mg reacts completely with O2
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Answer:

0.250 moles of MgO are produced when 0.250 mol of Mg reacts completely with O₂

Explanation:

In first place, the balanced reaction between Mg and O₂ is:

2 Mg + O₂ ⇒ 2 MgO

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reactants and products participate in the reaction:

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Then you can apply the following rule of three: if by reaction stoichiometry 2 moles of Mg produce 2 moles of MgO, 0.250 moles of Mg, how many moles of MgO will they form?

moles of MgO=\frac{0.250 moles of Mg*2 moles of MgO}{2 moles of Mg}

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