Answer :
The number of moles:
3.16g of [tex]KMnO_{4} [/tex]
1 mole of [tex]KMnO_{4} [/tex]- 158g.
x moles of [tex]KMnO_{4} [/tex]- 3.16g.
[tex]x=\frac{1\ mole*3.16g}{158g}=\frac{3.16}{158}=0.02\ mole[/tex]
The number of moles:
32.33g of [tex]K_{2}CrO_{4} [/tex]
1 mole of [tex]K_{2}CrO_{4} [/tex]- 194g.
x moles of [tex]K_{2}CrO_{4} [/tex]- 32.33g.
[tex]x=\frac{1\ mole*32.33g}{194g}=\frac{32.33}{194}=0.17\ mole[/tex]
The number of moles:
100g of [tex]KHCO_{3}[/tex]
1 mole of [tex]KHCO_{3}[/tex]- 100.1g.
x moles of [tex]KHCO_{3}[/tex]- 100g.
[tex]x=\frac{1\ mole*100g}{100.1g}=\frac{100}{100.1}=0.999\ mole[/tex]
Greetings,
n00nst00p :)
3.16g of [tex]KMnO_{4} [/tex]
1 mole of [tex]KMnO_{4} [/tex]- 158g.
x moles of [tex]KMnO_{4} [/tex]- 3.16g.
[tex]x=\frac{1\ mole*3.16g}{158g}=\frac{3.16}{158}=0.02\ mole[/tex]
The number of moles:
32.33g of [tex]K_{2}CrO_{4} [/tex]
1 mole of [tex]K_{2}CrO_{4} [/tex]- 194g.
x moles of [tex]K_{2}CrO_{4} [/tex]- 32.33g.
[tex]x=\frac{1\ mole*32.33g}{194g}=\frac{32.33}{194}=0.17\ mole[/tex]
The number of moles:
100g of [tex]KHCO_{3}[/tex]
1 mole of [tex]KHCO_{3}[/tex]- 100.1g.
x moles of [tex]KHCO_{3}[/tex]- 100g.
[tex]x=\frac{1\ mole*100g}{100.1g}=\frac{100}{100.1}=0.999\ mole[/tex]
Greetings,
n00nst00p :)
