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fiasKO [112]
3 years ago
15

An alkaline battery produces electrical energy according to the following equation.

Chemistry
1 answer:
VashaNatasha [74]3 years ago
4 0

a) Given reaction:

Zn + 2MnO2 + H2O → Zn(OH)2 + Mn2O3

1 mole of Zn combines with 2 moles of MnO2

Now:

# moles Zn present = 17.5/65.38 = 0.2677 moles

# moles of MnO2 present = 31.0/86.94 = 0.3566 moles

Since # moles of MnO2 is less than Zn i.e. it is not present in the 2:1 ratio (MnO2:Zn), MnO2 will be the limiting reagent

b) Based on stoichiometry:

2 moles of MnO2 produces 1 mole of Zn(OH)2

Thus, moles of Zn(OH)2 produced form the limiting reactant = 0.3566/2 = 0.1783 moles

Mass of Zn(OH)2 = 0.1783*99.42 = 17.73 g


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The  rate of a certain reaction is given by the following rate law:

            rate =  k [H_2][I_2]

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What is the reaction order in H_2?

What is the reaction order in I_2?

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At a certain concentration of H2 and I2, the initial rate of reaction is 2.0 x 104 M / s. What would the initial rate of the reaction be if the concentration of H2 were doubled? Round your answer to significant digits. The rate of the reaction is measured to be 52.0 M / s when [H2] = 1.8 M and [I2] = 0.82 M. Calculate the value of the rate constant. Round your answer to significant digits.

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The reaction order in H_2 is  n =  1

The reaction order in I_2 is  m = 1

The  overall reaction order z =  2

When the hydrogen is double the the initial rate is   rate_n  =  4.0*10^{-4} M/s

The rate constant is   k = 35.23 \  M^{-1} s^{-1}

Explanation:

From the question we are told that

   The rate law is  rate =  k [H_2][I_2]

   The rate of reaction is rate =  2.0 *10^{4} M /s

Let the reaction order for H_2 be  n and for I_2  be  m

From the given rate law the concentration of H_2 is raised to the power of 1 and this is same with I_2 so their reaction order is  n=m=1

   The overall reaction order is  

               z  = n +m

               z  =1 +1

               z  =2

At  rate =  2.0 *10^{4} M /s

        2.0*10^{4}  = k  [H_2] [I_2] ---(1)

= >    k  = \frac{2.0*10^{4}}{[H_2] [I_2]  }

given that the concentration of hydrogen is doubled we have that

            rate  = k [2H_2] [I_2] ----(2)

=>      k = \frac{rate_n  }{ [2H_2] [I_2]}

 So equating the two k

           \frac{2.0*10^{4}}{[H_2] [I_2]  } = \frac{rate_n  }{ [2H_2] [I_2]}

    =>    rate_n  =  4.0*10^{-4} M/s

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      rate_x =  52.0 M/s

        [H_2] = 1.8 M

         [I_2] =  0.82 \ M

We have

      52 .0 =  k(1.8)* (0.82)

     k = \frac{52 .0}{(1.8)* (0.82)}

      k = 35.23 M^{-2} s^{-1}

     

     

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