Stoichiometry

## Stoichiometry problems-

### Question 1- Balance the following chemical reactions:

#### Solution  – Balanced reactions are as follows,

a. 2CO + O2 2CO2

b. 2NaNO3 2NaNO2 + O2

c. 2O3 3O2

d. NH4NO3 N2O + 2H2O

e. 4CH3NH2 + 9O2 4CO2 + 10H2O + 2N2

f. Cr(OH)3 + 3HClO4 Cr(ClO4)3 + 3H2O

### Ans-balanced chemical equationsare as follows-

a) CaC2 + 2H2O —->  Ca(OH)2 + C2H2

b)  2KClO3    ——> 2KCl + 3O2

c)  2C6H6 + 15O2 —> 12CO2 + 6H2O

d) 2C5H12O  + 15O2 —-> 10CO2 + 12H2O

## Stoichiometry problems-

### Ans- Balanced equation is, 2Na2S2O3 + AgBr NaBr + Na3[Ag(S2O3)2] 2 mole          1 mole      1 mole

Part a) molar mass of AgBr = 187.77 gm/mole

mass of AgBr= 42.7 gm

moles= mass / molar mass = 42.7 / 187.77

moles of AgBr = 0.227 mole

1 mole AgBr is  needed to react completely with =    2 mole Na2S2O3

0.227 mole AgBr is  needed to react completely with =    2 x 0.227  mole Na2S2O3  = 0.454 mole

#### 0.454  moles of Na2S2O3 are needed to react completely with 42.7 g of AgBr.

Part b) molar mass of AgBr = 187.77 gm/mole

mass of AgBr= 42.7 gm

moles= mass / molar mass = 42.7 / 187.77

moles of AgBr = 0.227 mole

1 mole AgBr gives = 1 mole of NaBr

0.227 mole AgBr gives = 1 x 0.227 mole of NaBr

moles of NaBr = 0.227

molar mass of NaBr = 102.894 gm / mole

mass of NaBr = moles x molar mass = 0.227 x 102.894 = 23.3569

Mass of NaBr  that will be produced from 42.7 g of AgBr = 23.357 gm