Stoichiometry Avogadro's Number

Stoichiometry

Stoichiometry Using Avogadro's Number
Stoichiometry Avogadro's Number

This general chemistry video tutorial focuses on avogadro's number and how it's used to convert moles to atoms. This video also shows you how to calculate t. This unit is called the mole. A mole is the SI base unit for measuring the amount of a substance. One mole is 6.02. 10^23 particles. This number is called Avogadro's number, after Amedeo Avogadro. This quiz will cover the basics of counting small particles.You will need a calculator.

0.450 mole of Fe contains how many atoms? Solution: Start from the box labeled 'Moles of. We have to make a clear distinction between Avogadro's number and Avogadro's constant. The former is a unitless real number, defined as 6.02214076 × 10 23. The latter is a physical (or shall I call chemical?) constant, and thus has a unit. To appreciate the magnitude of Avogadro’s number, consider a mole of pennies. Stacked vertically, a mole of pennies would be 4.5 × 10 17 mi high, or almost six times the diameter of the Milky Way galaxy. If a mole of pennies were distributed equally among the entire population on Earth, each person would have more than one trillion dollars.

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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⁺²in 1.0 mole MgO = 6.022 x 10 ²³Mg⁺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

Stoichiometry Using Avogadro's Number

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