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Ganezh [65]
3 years ago
9

50cm3 of a mixture of methane and hydrogen were mixed with excess oxygen and exploded, the product after cooling to the original

room temperature were treated with excess koH solution when the volume decrease by 40cm3 , calculate the composition of the original mixture
Chemistry
1 answer:
marin [14]3 years ago
8 0

Answer:the initial composition of the reactants is

40cm^3 of CH4

40cm^3of H2

100cm^3 of H2O

Explanation:

Balanced reaction is

CH4 +H2+5/2O2______

CO2 +3H2O

Excess KOH at room temperature absorbs CO2 whose volume is given by 40cm^3 i.e the volume by which the solution decreases

So using Gay lussac combining ratio which states that gases combine in volumes that are in simple ratio to each other if gases.

Since CO2 in the equation is 1 mole

Means 1mole represent 40cm^3

So CH4:H2:O2 are in ratio of 1:1:5/2=(40:40:100)cm^3 respectively.

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Answer:

1.18×10²³ atoms.

Explanation:

From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02×10²³ atoms.

From the above concept, 1 mole of sodium also contains 6.02×10²³ atoms.

1 mole of sodium = 23 g.

Thus,

23 g of sodium contains 6.02×10²³ atoms.

Therefore, 4.5 g of sodium will contain = (4.5 × 6.02×10²³)/23 = 1.18×10²³ atoms.

From the above calculation,

4.5 g of sodium contains 1.18×10²³ atoms.

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