Question

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

Answer 1

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

Step-by-step explanation:

1. Balanced chemical equation (given)

`Cl_2+2KBr\rightarrow 2KCl+Br_2`

2. Mole ratios

`(1molCl_2)/(2molKBr)`

`(2molKCl)/(2molKBr)`

`(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

`(1molCl_2)/(2molKBr)* 0.092435molKBr=0.0462molCl_2`

`0.0462175molCl_2* 70.906g/molCl_2=3.28gCl_2`

b) KCl

`(2molKCl)/(2molKBr)*0.092435molKBr=0.0924molKCl`

`0.092435molKCl* 74.5513g/molKCl=6.89gKCl`

c) Br₂

`(1molBr_2)/(2molKBr)* 0.092435molKBr=0.0462molBr_2`

`0.0462175molBr_2* 159.808g/molBr_2=7.39gBr2`

The final calculations are rounded to 3 sginificant figures.