When \(0.2 \, \text{mol}\) of calcium reacts with \(880 \, \text{g}\) of water, \(2.24 \, \text{L}\) of hydrogen gas form at STP.
How would the amount of hydrogen produced change if the volume of water was decreased to \(440 \, \text{mL}\) (\(400 \, \text{g}\))?
Answer (3)
The volume of hydrogen produced would be the same.
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The amount of hydrogen produced does not change if the volume of water was decreased to 440mL 400g.
What is Mole?
The mole is an amount unit similar to familiar units like pair, dozen, gross, etc. It provides a specific measure of the number of **atoms or molecules **in a bulk sample of matter.
A mole is defined as the amount of substance containing the same number of atoms, molecules, ions, etc. as the number of atoms in a sample of pure 12C weighing exactly 12 g.
The reaction will be;
Ca(s) + 2 H₂O(l) → Ca(OH)₂(aq) + H₂ (g)
Since water is in excess, because 0.1 moles of **calcium **requires only 0.2 moles of water which is 3.6 g, therefore changing the mass of water used will not affect the amount of hydrogen gas produced since the amount of hydrogen gas produced depends on the amount of calcium used.
Therefore, The amount of hydrogen produced does not change if the volume of water was decreased to 440mL 400g.
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The amount of hydrogen gas produced remains the same at 2.24 L regardless of whether 880 g or 400 g of water is used because calcium is the limiting reactant in the reaction. As long as there is excess water, the volume of hydrogen produced will not change. Therefore, the decrease in water does not impact the amount of hydrogen gas generated from the reaction with calcium.
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