When sodium Metal is dropped into water, all the metal reacts in the equation shown below. What mass of hydrogen is liberated from a 46.0 gram piece of sodium? 2Na(s)+2H2O(I) > 2NaOH(aq)+H2(g)

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To calculate the mass of hydrogen liberated from a 46.0 gram piece of sodium, we need to determine the number of moles of sodium and then use the stoichiometry of the balanced equation to find the number of moles of hydrogen produced.

1. Convert the mass of sodium to moles:
We can use the molar mass of sodium to convert the mass to moles. The molar mass of sodium (Na) is 22.99 g/mol. So, we divide the given mass by the molar mass to get the number of moles:

Number of moles of sodium = 46.0 g / 22.99 g/mol

2. Use the stoichiometry of the balanced equation:
From the balanced equation, we can see that 2 moles of sodium react with 2 moles of water to produce 1 mole of hydrogen.
So, the mole ratio between sodium and hydrogen is 2:1.

3. Calculate the number of moles of hydrogen:
Since the mole ratio is 2:1, the number of moles of hydrogen produced will be half of the number of moles of sodium.

Number of moles of hydrogen = (Number of moles of sodium) / 2

4. Convert moles of hydrogen to mass:
To find the mass of hydrogen, we multiply the number of moles of hydrogen by the molar mass of hydrogen. The molar mass of hydrogen (H2) is 2.02 g/mol.

Mass of hydrogen = (Number of moles of hydrogen) * (Molar mass of hydrogen)

By following these steps, you can calculate the mass of hydrogen liberated from the given 46.0 gram piece of sodium.