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The alkali metal lithium reacted with water. Calculate how many grams, moles of hydrogen were released if 56 g of lithium reacted?​

User Dzhuneyt
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Lithium Reaction: Hydrogen Moles

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The alkali metal lithium reacted with water. Calculate how many grams, moles of hydrogen were released if 56 g of lithium reacted?

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To calculate the number of grams and moles of hydrogen released when 56 g of lithium reacts with water, we need to determine the balanced chemical equation for the reaction.

The balanced chemical equation for the reaction between lithium and water is:

2 Li + 2 H₂O -> 2 LiOH + H₂

From the equation, we can see that 2 moles of lithium react to produce 1 mole of hydrogen gas.

Now, let's calculate the moles of lithium in 56 g of lithium using its molar mass:

Molar mass of lithium (Li) = 6.94 g/mol

Moles of lithium = Mass of lithium / Molar mass of lithium

= 56 g / 6.94 g/mol

≈ 8.06 mol

Since 2 moles of lithium produce 1 mole of hydrogen, we can calculate the moles of hydrogen produced:

Moles of hydrogen = (Moles of lithium) / 2

= 8.06 mol / 2

= 4.03 mol

Finally, we can calculate the grams of hydrogen using the molar mass of hydrogen:

Molar mass of hydrogen (H₂) = 2.02 g/mol

Grams of hydrogen = Moles of hydrogen × Molar mass of hydrogen

= 4.03 mol × 2.02 g/mol

≈ 8.15 g

Therefore, when 56 g of lithium reacts with water, approximately 4.03 moles (or 8.15 grams) of hydrogen gas are released.

User Matanox
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