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zhuklara [117]
3 years ago
8

Ammonium perchlorate nh4clo4 is a powerful solid rocket fuel, used in the space shuttle boosters. it decomposes into nitrogen n2

gas, chlorine cl2 gas, oxygen o2 gas and water vapor, releasing a great deal of energy. calculate the moles of ammonium perchlorate needed to produce 0.10mol of chlorine. be sure your answer has a unit symbol, if necessary, and round it to the correct number of significant digits.
Chemistry
1 answer:
babymother [125]3 years ago
4 0
                              2NH4ClO4 --------> N2 + Cl2 + 2O2 + 4H2O
from reaction         2 mol                              1 mol
given                     x mol                              0.10 mol

Proportion:
<u>2 mol NH4ClO4 </u>= <u>1 mol Cl2</u>
x mol  NH4ClO4   0.10 mol Cl2

x= (2*0.10)/1 = 0.20 mol NH4ClO4
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Mass = 35g Volume = 7cm³ What is the Density?​
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rate of a certain reaction is given by the following rate law: rate Use this information to answer the questions below. What is
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Complete Question

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

            rate =  k [H_2][I_2]

rate Use this information to answer the questions below.

What is the reaction order in H_2?

What is the reaction order in I_2?

What is overall reaction order?

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.

Answer:

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

So when

      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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