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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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2Li + H2SO4=Li2SO4 + H2 How many liters of hydrogen gas, H2 at STP can be produced from 3.0 moles of Li? The molar volume of a g
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Answer:

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7 0
3 years ago
How much carbon would be needed to make 8.8g of carbon dioxide?<br>​
PIT_PIT [208]
The calculation for such a question can be achieved via Avogadro hypothesis

We know molar mass of CO2 is 44g/mole which is the sum of atomic masses i.e; C and 2 oxygen atoms

Molar mass of CO2 =12(C)+2*16(O) = 44 g/mole will contain 6.023 ※10^23 CO2 molecules ..

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=> 1g = (6.023/44) ※10^23 CO2 molecules

==> 8.80g = 8.80(6.023÷44)10^23 = 1.2046 ※10^23 molecules of CO2….

Thus there r 1.2046 ※10^23 molecules of CO2 in 8.80g

if u need to calculate no. of carbon atoms then multiply result by 1 and if u need no of oxygen atoms in 8.80g of co2 then multiply the result by 2 ….
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