Question

What is the reaction for CL^2 + 2 KBr —> 2 KCL+Br^2 of 11 grams of potassium bromide?

Answer 1

Answer:

All the amounts of reactants and products are:

KBr

          11.0g (given)

          0.0924 mol

Cl₂

           0.0462 mol

           3.28g

KCl

           0.0924 mol

           6.89 g

Br₂

         0.0462 mol

         7.39 g

         

Explanation:

1. Balanced chemical equation (given)

   $Cl_2+2KBr\rightarrow 2KCl+Br_2$

2. Mole ratios

     $\dfrac{1molCl_2}{2molKBr}$

      $\dfrac{2molKCl}{2molKBr}$

     $\dfrac{1molBr_2}{2molKBr}$

3. Molar masses

• Molar mass Cl₂: 70.906g/mol

• Molar mass KBr: 119.002 g/mol

• Molar mass KCl: 74.5513 g/mol

• Molar mass KBr: 159.808 g/mol

4. Convert 11 grams of potassium bromide to moles:

• #moles = mass in grams / molar mass

• #mol KBr = 11g / 119.002g/mol = 0.092435mol KBr

5. Use the mole ratios to find the amounts of Cl₂, KCl, and Br₂

a) Cl₂

       $\dfrac{1molCl_2}{2molKBr}\times 0.092435molKBr=0.0462molCl_2$

        $0.0462175molCl_2\times 70.906g/molCl_2=3.28gCl_2$

b) KCl

       $\dfrac{2molKCl}{2molKBr}\times0.092435molKBr=0.0924molKCl$

      $0.092435molKCl\times 74.5513g/molKCl=6.89gKCl$

c) Br₂

       

         $\dfrac{1molBr_2}{2molKBr}\times 0.092435molKBr=0.0462molBr_2$

         $0.0462175molBr_2\times 159.808g/molBr_2=7.39gBr2$

The final calculations are rounded to 3 sginificant figures.