For the balanced reaction 2 Na + 2 H2O -->2 NaOH + H2
how many grams of Hydrogen are produced if 120 g of Sodium are available?
Just follow the steps in this worked example.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To determine the amount of hydrogen produced, we need to use stoichiometry, which is the relationship between the amounts of reactants and products in a balanced chemical equation.
Here, the balanced equation shows that 2 moles of sodium (Na) react to produce 1 mole of hydrogen gas (H2). First, we need to convert the given mass of sodium (120 g) into moles.
To do this, we need the molar mass of sodium (Na), which is approximately 22.99 g/mol.
So, moles of sodium = given mass / molar mass
= 120 g / 22.99 g/mol
≈ 5.22 mol
Now, we can use the stoichiometry of the balanced equation to find the moles of hydrogen produced.
From the balanced equation, we can see that 2 moles of sodium (Na) produce 1 mole of hydrogen gas (H2).
So, moles of hydrogen = moles of sodium × (1 mole of H2 / 2 moles of Na)
= 5.22 mol × (1/2)
= 2.61 mol
Finally, to determine the mass of hydrogen produced, we multiply the moles of hydrogen by its molar mass, which is approximately 2.02 g/mol.
Mass of hydrogen = moles of hydrogen × molar mass
= 2.61 mol × 2.02 g/mol
≈ 5.28 g
Therefore, approximately 5.28 grams of hydrogen are produced when 120 grams of sodium are available for the given balanced reaction.