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)
[tex]Cl_2+2KBr\rightarrow 2KCl+Br_2[/tex]
2. Mole ratios
[tex]\dfrac{1molCl_2}{2molKBr}[/tex]
[tex]\dfrac{2molKCl}{2molKBr}[/tex]
[tex]\dfrac{1molBr_2}{2molKBr}[/tex]
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
[tex]\dfrac{1molCl_2}{2molKBr}\times 0.092435molKBr=0.0462molCl_2[/tex]
[tex]0.0462175molCl_2\times 70.906g/molCl_2=3.28gCl_2[/tex]
b) KCl
[tex]\dfrac{2molKCl}{2molKBr}\times0.092435molKBr=0.0924molKCl[/tex]
[tex]0.092435molKCl\times 74.5513g/molKCl=6.89gKCl[/tex]
c) Br₂
[tex]\dfrac{1molBr_2}{2molKBr}\times 0.092435molKBr=0.0462molBr_2[/tex]
[tex]0.0462175molBr_2\times 159.808g/molBr_2=7.39gBr2[/tex]
The final calculations are rounded to 3 sginificant figures.
