Stoichiometry

Stoichiometry

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Stoichiometry Numericals-

Question 1-

How many moles & atoms are in the following-

(i) 14.01 gm N (ii) 80.16 gm Ca (iii) 76.41 Vanadium

Solution-

 (i) 14.01 gm N

mass of N = 14.01 gm,molar mass of N = 14.01 gm/mole

moles = mass / molar mass

= 14.01 / 14.01

mole of N = 1 mole

1 mole of N has = 6.023x 1023 atoms

no. of N atoms = 6.023x 1023

 (ii) 80.16 gm Ca

mass of Ca = 80.16 gm,molar mass of Ca = 40.08 gm/mole

moles = mass / molar mass

= 80.16/ 40.08

mole of Ca = 2 mole

1 mole of Ca has = 6.023x 1023 atoms (avogadro’s no.)

2 mole of Ca has = 2  x 6.023x 1023 atoms

no. of Ca atoms = 12.046 x 1023

(iii) 76.41 gm Vanadium

mass of V = 76.41 gm,molar mass of  V = 50.94 gm/mole

moles = mass / molar mass

=76.41 / 50.94

mole of V = 1.5 mole

1 mole of V has = 6.023x 1023 atoms (avogadro’s no.)

1.5 mole of V has = 1.5  x 6.023x 1023 atoms

no. of  V atoms = 9.0345 x 1023

Question 2- How many Mg+2 ions are found in 1.00 mole of MgO ?

Ans-

1.0 mole MgO = 1.0 mole Mg+2

Therefore,

moles of  Mg+2  = 1.0

we know,

1.0 mole Mg+2   = Avogadro’s number

No. of  Mg+2   in 1.0 mole MgO =  6.022 x 10 23 Mg+2  ions

Stoichiometry Numericals-

Question 3- How many grams of Al can be created by decomposition of 9.8 gm of Al2O3 ?

Solution –

2Al2O3 —> 4Al        +  3O2

2 mole           4 mole       3 mole

mass of Al2O3 = 9.8 gm

molar mass of Al2O3= 102 gm/mole

moles of Al2O3 = mass / molar mass

= 9.8 / 102

moles of Al2O3 = 0.0961

2 moles Al2O3 gives = 4 mole Al

0.0961 moles Al2O3 gives = 4 x 0.0961 / 2 = 0.1922 mole 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

Mass of Al  = 5.19 gm

Question 4-

0.5 L of 0.20 M HCl solution is mixed with 0.5L of 0.40M AgNO3 solution & reaction occurs.If the reaction goes to completion ,what mass of AgCl produces ?

Ans-

HCl + AgNO3 —> AgCl + HNO3

Molarity of HCl = 0.20M

Volume of HCl = 0.5L

moles of HCl = Molarity x volume(L)

= 0.20 x 0.5

moles of HCl  = 0.1

Molarity of  AgNO3= 0.40M

Volume of AgNO3 = 0.5L

moles of AgNO3 = Molarity x volume(L)

= 0.40 x 0.5

moles of AgNO3 = 0.2

1mole HCl reacts with= 1 mole AgNO3

0.1mole HCl reacts with= 0.1 mole AgNO3

But 0.2 mole AgNO3 are present ,so HCl is limiting reagent.

0.1 mole HCl gives 0.1 mole AgCl.

molar mass of AgCl = 143.32 gm/mole

mass of AgCl = mole x molar mass = 0.1 x 143.32 = 14.332 gm

Mass of AgCl = 14.332 gm