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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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skad [1K]
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7 0
3 years ago
When 2.50 g of an unknown weak acid (ha) with a molar mass of 85.0 g/mol is dissolved in 250.0 g of water, the freezing point of
baherus [9]
When dT = Kf * molality * i
                = Kf*m*i
and when molality = (no of moles of solute) / Kg of solvent
                               = 2.5g /250g x 1 mol /85 g x1000g/kg
                               =0.1176 molal
and Kf for water = - 1.86 and dT = -0.255
by substitution 
0.255 = 1.86* 0.1176 * i
∴ i = 1.166
when the degree of dissociation formula is: when n=2 and  i = 1.166
a= i-1/n-1 = (1.166-1)/(2-1) = 0.359 by substitution by a and c(molality) in K formula
∴K = Ca^2/(1-a)
     = (0.1176 * 0.359)^2 / (1-0.359)
     = 2.8x10^-3



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For the reaction Na2CO3+Ca(NO3)2⟶CaCO3+2NaNO3 how many grams of calcium carbonate, CaCO3, are produced from 79.3 g of sodium car
Alexus [3.1K]

Answer:

74.81 grams of calcium carbonate are produced from 79.3 g of sodium carbonate.

Explanation:

The balanced reaction is:

Na₂CO₃ + Ca(NO₃)₂ ⟶ CaCO₃ + 2 NaNO₃

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

  • Na₂CO₃: 1 mole
  • Ca(NO₃)₂: 1 mole
  • CaCO₃: 1 mole
  • NaNO₃: 2 mole

Being the molar mass of the compounds:

  • Na₂CO₃: 106 g/mole
  • Ca(NO₃)₂: 164 g/mole  
  • CaCO₃: 100 g/mole
  • NaNO₃: 85 g/mole

then by stoichiometry the following quantities of mass participate in the reaction:

  • Na₂CO₃: 1 mole* 106 g/mole= 106 g
  • Ca(NO₃)₂: 1 mole* 164 g/mole= 164 g
  • CaCO₃: 1 mole* 100 g/mole= 100 g
  • NaNO₃: 2 mole* 85 g/mole= 170 g

You can apply the following rule of three: if by stoichiometry 106 grams of Na₂CO₃ produce 100 grams of  CaCO₃, 79.3 grams of Na₂CO₃ produce how much mass of  CaCO₃?

mass of CaCO_{3} =\frac{79.3 grams of Na_{2} CO_{3} *100 grams of of CaCO_{3}}{106 grams of Na_{2} CO_{3}}

mass of CaCO₃= 74.81 grams

<u><em>74.81 grams of calcium carbonate are produced from 79.3 g of sodium carbonate.</em></u>

6 0
3 years ago
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