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kirza4 [7]
3 years ago
14

What is the percent yield of hbr if 85.00 g of hbr was formed from 30. g of h2?

Chemistry
2 answers:
Dima020 [189]3 years ago
7 0
1) Chemical equation

H2 (g) + Br2 (g) ---> 2 HBr

2) molar ratios

1 mol H2 : 1 mol Br2 : 2 moles HBr

3) convert 30.0 g of H2 to number of moles

moles = mass in grams / molar mass = 30.0 g / 2.02 g/mol = 14.85 moles

4) calculate theoretical yield

2 moles HBr / 1 mol H2 = x / 14.85 moles H2

=> x = 14.85 moles H2 * 2 moles HBr / 1 mol H2 = 29.7 moles HBr

5) convert 59.52 moles HBr to mass

mass = number of moles * molar mass = 29.7 moles * 80.91 g /mol = 2403.03 g

6) percent yield = (actual yield / theoretical yield) * 100 = (85.00g / 2403.03g)*100 = 3.54%

Answer: 3.54%
vitfil [10]3 years ago
3 0

Ok, lets see the definitions of percent yield, actual yield and theoretical yield.

Percent yield is the ratio of actual yield to theoretical yield.

Actual yield- amount of product produced in the EXPERIMENT.

Theoretical yield- max amount of product produced through CALCULATIONS

% yield= actual yield (from experiment)/theoretical yield (from calculation) *100

1st Step: Write the reaction 

H2 + Br2 --> 2 HBr

2nd Step: Get the mass ratio of H2 and HBr to find theoretical yield

1 mole H2 gives 2 moles HBr ( molar mass of H2= 2 g/mol, HBr= 81 g/mol)

2 g H2 gives 2*81= 162 g HBr

30 g H2 gives 162*30/2 = 2430 g HBr ( The equation is 2 g H2/ 30 gH2= 162 g HBr/ x g HBr)

So theoretically 2430 g HBr are produced by calculation ( THEORETICAL YIELD)

By experiment 85 g HBr are produced as it is given at the question ( ACTUAL YIELD)

% yield= actual yield / theoretical yield *100 = 85/2430 *100= 3.5 %

The percent yield is 3.5 %.

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Why and how do ions form?
Zanzabum

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A chemist fills a reaction vessel with 9.20 atm nitrogen monoxide (NO) gas, 9.15 atm chlorine (CI) gas, and 7.70 atm nitrosyl ch
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Answer:

The reactions free energy \Delta G = -49.36 kJ

Explanation:

From the question we are told that

      The pressure of (NO) is P_{NO} = 9.20 \ atm

      The  pressure of  (Cl) gas is  P_{Cl} = 9.15 \ atm

       The  pressure of nitrosly chloride (NOCl) is P_{(NOCl)} = 7.70 \ atm

The reaction is

              2NO_{(g)} + Cl_2 (g)    ⇆   2 NOCl_{(g)}

 From the reaction we can  mathematically evaluate the \Delta G^o (Standard state  free energy ) as

                    \Delta G^o = 2 \Delta G^o _{NOCl} -   \Delta G^o _{Cl_2}  - 2 \Delta G^o _{NO}

The Standard state  free energy for NO is  constant with a value  

                 \Delta G^o _{NO} = 86.55 kJ/mol

 The Standard state  free energy for Cl_2 is  constant with a value                  

             \Delta G^o _{Cl_2} = 0kJ/mol

 The Standard state  free energy for NOCl is  constant with a value

         \Delta G^o _{NOCl} =66.1kJ/mol

Now substituting this into the equation

        \Delta G^o = 2 * 66.1 - 0 - 2 * 87.6

                = -43 kJ/mol

The pressure constant is evaluated as

         Q =  \frac{Pressure \ of  \ product }{ Pressure  \ of \ reactant }

Substituting  values  

        Q = \frac{(7.7)^2 }{(9.2)^2 (9.15) } = \frac{59.29}{774.456}

           = 0.0765

The free energy for this reaction is evaluated as

           \Delta  G  =  \Delta  G^o  + RT ln Q

Where R is gas constant with a value  of  R = 8.314 J / K \cdot mol

          T is temperature in K  with a given value of  T = 25+273 = 298 K

   Substituting value

                \Delta  G  = -43 *10^{3} + 8.314 *298 * ln [0.0765]

                       = -43-6.36

                      \Delta G = -49.36 kJ

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uranmaximum [27]

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Ans: Potassium (K)

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