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n200080 [17]
3 years ago
5

Methane gas and chlorine gas react to form hydrogen chloride gas and carbon tetrachloride gas. What volume of hydrogen chloride

would be produced by this reaction if of methane were consumed? Also, be sure your answer has a unit symbol, and
Chemistry
1 answer:
IrinaK [193]3 years ago
5 0

The question is incomplete, here is the complete question:

Methane gas and chlorine gas react to form hydrogen chloride gas and carbon tetrachloride gas. What volume of hydrogen chloride would be produced by this reaction if 1.1 mL of methane were consumed? Be sure your answer has the correct number of significant digits.

<u>Answer:</u> The volume of hydrogen chloride produced in the reaction will be 4.4 mL

<u>Explanation:</u>

We are given:

Volume of methane gas = 1.1 mL

The chemical equation for the reaction of methane gas and chlorine gas follows:

CH_4(g)+4Cl_2(g)\rightarrow 4HCl(g)+CCl_4(g)

Moles of methane gas = 1 mole

Moles of hydrogen chloride gas = 4 moles

The relationship of number of moles and volume at constant temperature and pressure was given by Avogadro's law. This law states that volume is directly proportional to number of moles at constant temperature and pressure.

The equation used to calculate number of moles is given by:

\frac{V_1}{n_1}=\frac{V_2}{n_2}

where,

V_1\text{ and }n_1 are the volume and number of moles of methane gas

V_2\text{ and }n_2 are the volume and number of moles of hydrogen chloride

We are given:

V_1=1.1mL\\n_1=1mol\\V_2=?L\\n_2=4mol

Putting values in above equation, we get:

\frac{1.1}{1}=\frac{V_2}{4}\\\\V_2=\frac{1.1\times 4}{1}=4.4mL

Hence, the volume of hydrogen chloride produced in the reaction will be 4.4 mL

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A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ?C. The initial concentrations of Pb2+ and Cu2+ are
VARVARA [1.3K]

Answer:

a) Ecell = 0.5123 V

b) Ecell =  0.4695 V

c) [Pb2 +] = 4.75 M

Explanation:

a)

The reaction at the cathode is represented as follows:

Cu2 + + 2e- -> Cu (s) Eocathode = 0.34 V

The reaction at the anode is equal to:

Pb (s) -> Pb2 + + 2e- Eoanode = -0.13 V

The number of moles of the electrons that are involved is equal to n = 2

Standard cell potential equals Eo = Eocathode - Eoanode = 0.34 V- (-0.13 V) = 0.47 V

 The initial cell potential can be calculated with the following formula:

Ecell = Eocell - - 0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (0.052 / 1.4) = 0.5123 V

b)

The reaction in the cell is equal to:

Cu2 + + Pb (s) -> Cu (s) + Pb2 +

The concentration of Cu2 that gives the exercise is equal 0.2 M

Therefore, the change in concentration for Cu2 + is equal to:

Cu2 + = 1.4 M - 0.2 M = 1.2 M

We use the formula from part a)

Ecell = Eocell - (0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (1,252 / 1.2) = 0.4695 V

c)

To find the concentration of Pb2 + when there is a potential change in the cell of 0.37 V, we must clear the concentration of Pb2 + from the following formula:

Eccell = Echocell - (0.0592 / n) log (([Pb2 +]) / ([Cu2 +]))

0.0296 log ([Pb2 +] / [Cu2 +]) = (Eocélula - Ecélula / 0.0296)

Clearing Pb2 +:

[Pb2 +] = 4.75 M

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