Let, wt, of metal oxide be 100 g
wt of metal= 60 g
wt of oxygen= (100-60) g= 40 g

Eq. wt. of metal = wt. of metal/ wt of oxygen* 8
= 12
As metal is divalent, At. wt= Eq. wt* valency= 12*2 = 24

Ans:
In this reaction, the total mass of reactants is equal to the total mass of the product.

This illustrates the law of conservation of mass which states, "Under similar conditions of temperature and pressure, the total mass of reactants in any chemical reaction is equal to the total mass of product."

Ans:Solution:
Given,
Experiment 1,

Wt. of zinc = 0.936 g
Wt. of oxide = 1.165 g
Wt. of oxygen = (1.165 – 0.936) g = 0.229 g

The ratio of weight of oxygen is to zinc in zinc oxide is

Experiment 2,

Wt. of zinc = 1.236 g
Wt. of oxide = 1.538 g
Wt. of oxygen = (1.538 – 1.236) g = 0.302 g

The ratio of weight of oxygen is to zinc in this oxide is

The ratio of weight of oxygen to weight of zinc in both the oxides is the same. Hence, given data
illustrate the law of definite proportions.

Alternative Method

In...

Here,
In P₂O₃, 62gm of Phosphorous combined with 48gm of oxygen.
In P₂O₅, 62gm of Phosphorous combined with 80gm of oxygen.

Ratio of mass of oxygen in P₂O₃and P₂O₅= 48:80 = 3:5 which is simple whole number ratio.

In this reaction, the ratio of mass of oxygen in PO and PO combined with fixed mass of phosphorous bear simple whole number ratio.

This illustrates, the law of multiple proportion of stoichiometry which states that,"Whentwo elements combined to each other to give two or more productsthen...

Solution:
Given,
In the compound X,
% of hydrogen = 25% & % of carbon = (100 – 25)% = 75% 25 parts by weight of hydrogen combine with 75 parts by weight of carbon.
1 part by weight of hydrogen combine withparts by weight of carbon
= 3 parts by weight of carbon

In the compound Y,
% of hydrogen = 14.3% & % of carbon = (100 – 14.3)% = 85.7% 14.3 parts by weight of hydrogen combine with 85.7 parts by weight of carbon
1 part by weight of hydrogen combine withparts by weight of carbon
=5.99 parts by...

Nitrogen combines with oxygen to form various oxide.

Compound	Combining	element	Ratio of	masses	with fixed	mass of N
N2O3	N	O	28	48	14	24
N2O5	N	O	28	80	14	40

The ratio of different weight of oxygen 24, 40 combining with fixed weight of nitrogen 14 is 24:40 i.e. 3:5 which is simple whole number ratio. The given data illustrate the law of multiple proportions.

Statement:

"When two elements combine to form more than one compound, then the different weight of one element combining with a...

Law of reciprocal proportion relates the proportions in which elements combine across a number of different elements. It is also called as equivalent proportion.

Example:

If element A combines with element B and also with C, if B and C combine together, the proportion by weight in which they do so will be simply related to the weights of B and C which separately combine with a constant weight of A.

Statement:

A chemical compound always contains the same elements combined together in the fixed ratio by weight, irrespective of the origin.

According to the Dalton's atomic theory, law of conservation of mass states that, "Matter is made up if extremely small indivisible particles which can neither be created nor be destroyed."

Under similar condition of temperature and pressure, gaseous reactants and products combine in a simple whole number ratio of their volumes.

When two different elements combine separately with the fixed weight of the elements, the ratio in which they do so will be the same or some simple multiple of the ratio which they combine together.

C + 2H₂ CH₄

Total weight of reactant = 16g
Total weight of product = 16g
Total weight of reactant = total weight of product

So these data show the law of conservation of mass.

Statement:

Total sum of mass and energy of system remains constant, but they can be interconverted.

The law of conservation of mass is known as indestructibility of matter because there is no loss of mass when reactant reacts to form product i.e. total mass of the reactant is equal to total mass of the product.

In 1803, John Dalton proposed the law of multiple proportion. Law of multiple proportion states that, "When two elements which combine to form two or more different compounds, then the weight of one of the elements which combine with the constant weight of the other bear simple whole number ratio."

Given,

No. of atoms of an element = 2 x 10²¹
Mass of atoms of an element = 0.4g

According to the questions,

2 x 10²¹ atoms of an element has mass 0.4g
6.022 x 1023 atoms of element has mass = 120.44g

i.e. 1 mole of atom of an element has mass 120.44g

Hence, the mass of 0.5 mole of an elements is 0.5 x 120.44g = 60.22g

Given,

The given weight of CO = 0.28 g
The molecular weight of CO = 28g

No. of moles of CO = 0.01 mole

We know that,

1 mole of CO contains 6.022 x 10²³ molecules
0.01 mole of CO contains 0.01 x 6.022 x 10²³ molecules = 6.022 x 10²¹ molecules

The number of molecules left is (6.022 x 10²¹ - 2 x 10²¹) = 4.022 x 10²¹ molecules

We know that,

6.022 x 10²³ molecules of CO = 1 mole of CO

4.012 x 10²¹ molecules of CO = 6.68 x 10^-3

Hence, no. of mole of CO left is 6.68 x 10^-3

Two atoms of Carbon:

1 mole of carbon = 12g
1 mole of carbon = 6.022 x 10²³ carbon atom

Thus,
6.022 x 10²³ carbon atom has mass 12g
2 carbon atom has mass = 3.98 x 10^-23 g carbon.

Three molecules of hydrogen:

1 mole of hydrogen molecule = 2g
1 mole of hydrogen molecule = 6.022 x 10²³ hydrogen molecule

Thus,
6.022 x 10²³ hydrogen molecule has mass 2g hydrogen.

3 hydrogen molecules has mass = 9.96 x 10^-24g

The balanced chemical equation is given as:

S + O₂ SO₂

According to the reaction

6.022 x 10²³ molecules of SO₂ are produced from 32g sulphur.

6.022 x 10²⁴ molecules of SO₂ are produced from = 320g sulphur

Similarly,

6.022 x 10²³ molecules of SO₂ are produced from 32g of oxygen

6.022 x 10²⁴ molecules of SO₂ can be produced from = 320g oxygen

The balanced chemical equation is given as

C + O₂ CO₂

Given mass of CO₂ = 220mg = 0.22g

44g of CO₂ is produced from 6.022 x 10²³ oxygen molecules.

0.22g of CO₂ can be produced from 3.011 x 10²¹ oxygen molecules.

Stoichiometry is defined as the branch of chemistry that deals with the mass-volume relationship of elements in a chemical change.

The etymological meaning of stoichiometry is measurement of elements. In Greek, stoicheion means element and metron means to measure. Stoichiometry = Stoicheion + Metron

Those substances that obey the laws of Stoichiometry are called stoichiometric compounds.

The difference between Gram atomic mass and Gram atomic molecular are given below:

Gram atomic mass	Gram molecular mass
1.The mass of 6.023 x 1023 atoms of one mole atom of any element when expressed in terms of grams is called gram atomic mass.	1.The mass of 6.023 x 1023 molecules or 1 mole molecule of any substance expressed in grams is called gram molecular mass.
2.It is used to calculate the mass of an atom.	2. It is used to calculate the mass of a molecule/molecular compound.
3....

One mole is the SI unit for the amount of substance or particles. The particles can be atoms, molecules or ions.Thus, it is the fixed collection of 6.023 x 10²³number of atoms, molecules, ions, electrons, protons, etc particles.

The fixed number6.023 x 10²³is called the Avogadro's number.

Number of moles can be expressed in terms of mass, volume and number of particles as:

Solution:
The required balance chemical equation:

2Mg + O₂---------------> 2MgO

i) For limiting reactant.
According to the balanced chemical equation.

48g Mg reacts with 32g of O₂
1g Mg reacts with (32 48)g of O₂
2g Mg reacts with 0.67 2g O₂

But the mass of O₂given is 3g. This means that Mg is finished first in the reaction. So, Mg is limiting reactant.

ii.
Given mass of O₂= 3g
Consumed mass of O₂= 1.33g

Therefore,
No. of O₂left over =...