A gas of certain mass occupies a volume of 650cubic centimeterunder a pressure of 750mm Hg.calculate the pressure under which th
e volume of the gas will be reduced by 10 percent of its original volume. Solve it using Boyle's law of equation.
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Answer: The volume of hydrogen gas produced will be, 12.4 L
Explanation : Given,
Mass of = 54.219 g
Number of atoms of =
Molar mass of = 98 g/mol
First we have to calculate the moles of and
.
and,
Now we have to calculate the limiting and excess reagent.
The balanced chemical equation is:
From the balanced reaction we conclude that
As, 3 mole of react with 2 mole of
So, 0.553 moles of react with
moles of
From this we conclude that, is an excess reagent because the given moles are greater than the required moles and
is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of
From the reaction, we conclude that
As, 3 mole of react to give 3 mole of
So, 0.553 mole of react to give 0.553 mole of
Now we have to calculate the volume of gas at STP.
As we know that, 1 mole of substance occupies 22.4 L volume of gas.
As, 1 mole of hydrogen gas occupies 22.4 L volume of hydrogen gas
So, 0.553 mole of hydrogen gas occupies volume of hydrogen gas
Therefore, the volume of hydrogen gas produced will be, 12.4 L