Stoichiometry

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Stoichiometry Numericals-

Question 1- How many moles of AlCl3 are required to react with 4.5 gm of Ca ?

3Ca+ 2AlCl3 —> 3CaCl2 + 2 Al

Solution –

mass of Ca = 4.5 gm , molar mass of Ca= 40.08 gm/mole

moles of Ca = mass / molar mass = 4.5 / 40.08

moles of Ca = 0.112

3 moles Ca reacts with = 2 moles AlCl3

0.112 moles Ca reacts with = 2 x 0.112 / 3  moles AlCl3

Moles of AlCl3 are required to react with 4.5 gm of Ca = 0.0746

Question 2- How many moles are in 7.15 gm H2 ?

Solution- mass of H2 = 7.15 gm

molar mass of H2 = 2 gm / mole

no. of moles = mass / molar mass

= 7.15 / 2 = 3.575

no. of moles = 3.58

3.58 moles are in 7.15 gm H2. Ans

Question 3-  Calculate the no. of moles in 8.75 gm C6H12O6 .

Solution –

mass of C6H12O6 = 8.75 gm

molar mass of C6H12O6 (glucose)  = 180.16 gm / mole

no. of moles = mass / molar mass

= 8.75 / 180.16 = 0.0486

no. of moles = 0.0486

no. of moles in 8.75 gm C6H12O6 = 0.0486

Stoichiometry Numericals-

Question 4- How many moles are in 100 gm of H2O ?

Solution- mass of H2O = 100 gm

molar mass of H2O = 18 gm / mole

no. of moles = mass / molar mass

= 100 / 18 = 5.56

no. of moles = 5.56

5.56 moles are in 100 gm of H2O. Ans

Question 5- Find number of  moles in 15.60 gm of CH4?

Solution- mass of CH4 = 15.60 gm

molar mass of CH4 = 16 gm / mole

no. of moles = mass / molar mass

= 15.60 / 16 = 0.975

no. of moles = 0.975

0.975 moles are present in 15.60 gm of CH4. Ans

Question 6- If a reaction  starts with 10.0 moles of propane, how many grams of CO2 will it yield?

Solution –

C3H8 + 5O2 —> 3CO2 + 4H2O

1.0 mole propane on combustion gives = 3 moles CO2

10.0 mole propane on combustion gives = 3 x 10 moles CO2 = 30 moles

moles of CO2 = 30.0

molar mass of CO2 = 44.01 gm / mole

mass of CO2 = moles x molar mass = 30.0 x 44.01 = 1320.3 gm

Mass of CO2 = 1320.3 gm. ans

Question 7- How many grams of Al can be obtained by the decomposition of 9.8 gm Al2O3 ?

Solution-

2Al2O3 —-> 4 Al + 3O2

mass of Al2O3 = 9.8 gm

molar mass of Al2O3 = 101.96 gm / mole

no. of moles of Al2O3 = mass / molar mass

= 9.8 / 101.96 = 0.0961

no. of moles of Al2O3 = 0.0961

2.0 mole Al2O3  gives = 4 moles Al

0.0961 mole Al2O3 gives = 4 x 0.0961 / 2.0 moles Al

moles of Al = 0.1922

molar mass of Al = 27 gm / mole

mass of Al = moles x molar mass = 0.1922 x 27 = 5.1894 gm

Mass of Al = 5.189 gm. ans