What is the reaction for CL^2 + 2 KBr —> 2 KCL+Br^2 of 11 grams of potassium bromide?
Answer choices:
1 mol
159.808g
7.39g
119.002g
2
2mol
11.0g
Br^2
KBr

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)

2. Mole ratios

3. Molar masses

4. Convert 11 grams of potassium bromide to moles:

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

a) Cl₂

b) KCl

c) Br₂

The final calculations are rounded to 3 sginificant figures.

the answer is 303.15 or 303

i think it's c. if you look search electronic symbol on google and then click on wiki it shows you a