The volume of hydrogen which is required to produce 11.2 L of water vapour (at STP)=
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Answer:
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For the production of water, the following reaction will take place between hydrogen gas a and oxygen gas
2H₂ + O₂ → 2H₂O
According to this equation, 2 moles of H₂ will react with 1 mole of O₂ to produce 2 moles of H₂O.
We know that 1 mole contains 22.4L of an ideal gas at STP
Let us consider these gases as ideal
2H₂ + O₂ → 2H₂O
2x22.4L 22.4L 2x22.4L
On diving the equation by 2
H₂ + 1/2O₂ → H₂O
22.4L 11.2L 22.4L
On further dividing the equation by 2
1/2H₂ + 1/4O₂ →1/2 H₂O
11.2L 5.6L 11.2L
Therefore, we can see that 11.2L of hydrogen gas will produce 11.2L of water.