Stoichiometry problems- source: njit.edu

## Stoichiometry problems-

### Question 1 –

#### Part a) Balanced equation is ,

4KO2 + 2H2O 3O2       + 4KOH

4 mole    2 mole      3 mole     4 mole

mass of O2 formed= 235 gm

molar mass of O2 = 32 gm/mole

moles of O2 = mass / molar mass = 235/32 = 7.344

3 mole O2 is produced from = 4 mole KO2

7.344 mole O2 is produced from = 4 x 7.344 / 3 = 9.792 mole KO2

moles of KO2 = 9.792

molar mass of KO2 = 71 gm/mole

mass of KO2 = moles x molar mass = 9.792 x 71

695.232 gm mass of KO2 produces 235 g of O2

### Part b)

4KO2 + 2H2O 3O2       + 4KOH

mass of KO2= 123 gm ; molar mass of KO2 = 71 gm/mole

moles of KO2 = mass / molar mass = 123 / 71

moles of KO2  = 1.73

According to reaction ,

moles of KO2 = moles of KOH

therefore moles of KOH = 1.73

KOH + CO2 KHCO3

1 mole KOH removes one mole of CO2, thus 1.73 moles of KOH removes 1.73 mole of CO2.

Therefore moles of CO2 = 1.73

molar mass of CO2 = 44 gm / mole

mass of CO2 = moles x molar mass = 1.73 x 44 = 76.12 gm

76.12 gm mass of CO2 can be removed by 123 g of KO2.

## Stoichiometry problem-

### Question 2-How much Calcium is present in Ca(NO3)2 that contains 20 gm nitrogen?

#### Solution –

In calcium nitrate, Ca(NO3)2, two atoms of nitrogen are present for every one atom of calcium.

The atomic  mass of nitrogen = 14

The atomic  mass of calcium = 40

The molar mass of Ca(NO3)2 = 40 + 2(14 + 3 x 16)= 164 gm /mole

28 gm nitrogen is present in calcium nitrate, when Ca is = 40 gm

20 gm nitrogen is present in calcium nitrate, when Ca is = 40 x 20 / 28  gm= 28.57 gm