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Pavel [41]
3 years ago
7

Solution stoichiometry allows chemists to determine the volume of reactants or products involved in a chemical reaction when the

chemical substances are dissolved in water. Instead of using mass to calculate the resulting amount of reactants or products, molarity and volume will be used to calculate the moles of substances. A 20.00 mL sample of a 0.250 M solution of HCl reacts with excess Ba(OH)2. What mass of H2O is produced in the reaction?
Chemistry
1 answer:
andrey2020 [161]3 years ago
8 0

Answer:

90 mg of H₂O

Explanation:

The reaction that takes place is:

2HCl + Ba(OH)₂ → BaCl₂(aq) + 2H₂O

With the information given by the problem and the definition of <em>molarity</em> (M=n/V), we can calculate the moles of HCl:

20.00 mL * 0.250 M = 5 mmol HCl

Now we use the <em>stoichiometric ratio</em> to <u>convert moles of HCl</u> to moles of H₂O and then to mass of H₂O:

5 mmol HCl * \frac{2mmolH_{2}O}{2mmolHCl} *\frac{18mg}{1mmolH_{2}O} = 90 mg H₂O

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In thermodynamics, we determine the spontaneity of a reaction by the sign of ΔG. In electrochemistry, spontaneity is determined
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<u>Answer:</u>

<u>For A:</u> The standard cell potential of the reaction is 4.4 V

<u>For B:</u> The standard Gibbs free energy of the reaction is -8.50\times 10^5J

<u>For C:</u> The reaction is spontaneous as written.

<u>Explanation:</u>

  • <u>For A:</u>

The given chemical reaction follows:

2Li(s)+Cl_2(g)\rightarrow 2Li^+(aq.)+2Cl^-(aq.)

The given half reaction follows:

<u>Oxidation half reaction:</u>  Li(s)\rightarrow Li^+(aq.)+e^-;E^o_{Li^+/Li}=-3.04V ( × 2)

<u>Reduction half reaction:</u>  Cl_2(g)+2e^-\rightarrow 2Cl^-(aq.);E^o_{Cl_2/2Cl^-}=+1.36V

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction.

Here, chlorine will undergo reduction reaction will get reduced.

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

E^o_{cell}=1.36-(-3.04)=4.4V

Hence, the standard cell potential of the reaction is 4.4 V

  • <u>For B:</u>

Relationship between standard Gibbs free energy and standard electrode potential follows:

\Delta G^o=-nFE^o_{cell}

where,

n = number of electrons transferred = 2mol\text{ e}^-

F = Faradays constant = 96500J/V.mol\text{ e}^-

E^o_{cell} = standard cell potential = 4.4 V

Putting values in above equation, we get:

\Delta G^o=-2\times 96500\times 4.4=-849200J=-8.50\times 10^5J

Hence, the standard Gibbs free energy of the reaction is -8.50\times 10^5J

  • <u>For C:</u>

For a reaction to be spontaneous, the standard Gibbs free energy change of the reaction must be negative.

From above, the standard Gibbs free energy change of the reaction is coming out to be negative.

Hence, the reaction is spontaneous as written.

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