Calculate the volume of hydrogen, in L, produced when 12.15 g of magnesium react completely with water. Assume S..T.P.

Mg + H2O ==> MgO + H2

mols Mg = 12.15/atomic mass Mg
Convert mols Mg to mols H2 using the coefficients in the balanced equation.
Now convert mols H2 to L remembering that 1 mol of any gas occupies 22.4L at STP

To calculate the volume of hydrogen gas produced when magnesium reacts with water, we need to make use of the balanced chemical equation for the reaction between magnesium and water, which is:

Mg + 2H₂O → Mg(OH)₂ + H₂

From the balanced equation, we can see that one mole of magnesium reacts to produce one mole of hydrogen gas.

Step 1: Calculate the number of moles of magnesium:
To determine the number of moles, we use the formula:

moles = mass / molar mass

In this case, the molar mass of magnesium (Mg) is 24.31 g/mol.

moles of Mg = 12.15 g / 24.31 g/mol = 0.500 moles

Step 2: Calculate the volume of hydrogen gas:
Since the reaction takes place at standard temperature and pressure (STP), we can use the molar volume of a gas at STP, which is 22.4 L/mol.

volume of H₂ = moles of H₂ * molar volume
volume of H₂ = 0.500 moles * 22.4 L/mol = 11.2 L

Therefore, the volume of hydrogen produced when 12.15 g of magnesium react completely with water at STP is 11.2 L.