Question

Calculate the number of moles 3.16 g of KMnO4 32.33 g of K2CrO4 100 g of KHCO3

Answer 1

The number of moles:
3.16g of $KMnO_{4}$
1 mole of $KMnO_{4}$- 158g.
x moles of $KMnO_{4}$- 3.16g.
$x=\frac{1\ mole*3.16g}{158g}=\frac{3.16}{158}=0.02\ mole$

The number of moles:
32.33g of $K_{2}CrO_{4}$
1 mole of $K_{2}CrO_{4}$- 194g.
x moles of $K_{2}CrO_{4}$- 32.33g.
$x=\frac{1\ mole*32.33g}{194g}=\frac{32.33}{194}=0.17\ mole$

The number of moles:
100g of $KHCO_{3}$
1 mole of $KHCO_{3}$- 100.1g.
x moles of $KHCO_{3}$- 100g.
$x=\frac{1\ mole*100g}{100.1g}=\frac{100}{100.1}=0.999\ mole$


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