Stoichiometry

stoichiometry

Stoichiometry problems-

Question 1- Balance the following chemical reactions:

a. CO + O2 CO2

b. NaNO3 NaNO2 + O2

c. O3 O2

d. NH4NO3 N2O + H2O

e. CH3NH2 + O2 CO2 + H2O + N2

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

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

Question 2- Write the balanced chemical equations –

a. Calcium carbide (CaC2) reacts with water to form calcium hydroxide (Ca(OH)2) and acetylene gas (C2H2).

b. When potassium chlorate (KClO3) is heated, it decomposes to form KCl and oxygen gas (O2).

c.  Combustion of benzene in air.

d. Combustion of C5H12O

Ans- balanced chemical equations are 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-

Question 3-  Given the following reaction:

Na2S2O3 + AgBr NaBr + Na3[Ag(S2O3)2]

a. How many moles of Na2S2O3 are needed to react completely with 42.7 g of AgBr?

b. What is the mass of NaBr that will be produced from 42.7 g of AgBr?

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