How many grams of magnesium metal will react completely with 5.2 liters of 4.0 M HCl? Show all of the work needed to solve this problem.
Mg (s) + 2HCl (aq) MgCl2 (aq) + H2 (g)
mols HCl you have = M x L = ?
mols Mg needed = 1/2 mols HCl because you know 1 mol Mg reacts with 2 mols HCl.
grams Mg = mols Mg x atomic mass Mg.
To solve this problem, we can use the concept of stoichiometry, which allows us to relate the amount of one substance to another in a balanced chemical equation.
First, let's determine the amount of moles of HCl in 5.2 liters of 4.0 M HCl.
Molarity (M) is defined as moles of solute per liter of solution. Therefore, 4.0 M HCl means there are 4.0 moles of HCl dissolved in 1 liter of solution.
To find the moles of HCl in 5.2 liters of 4.0 M HCl, we can use the following equation:
moles of HCl = molarity × volume
moles of HCl = 4.0 M × 5.2 L = 20.8 moles of HCl
Now, based on the balanced chemical equation, we know that the ratio between moles of magnesium (Mg) and moles of HCl is 1:2. This means that for every 1 mole of magnesium, 2 moles of HCl will be consumed or reacted.
Therefore, the number of moles of magnesium that will react with 20.8 moles of HCl is half that amount:
moles of Mg = 1/2 × moles of HCl
moles of Mg = 1/2 × 20.8 moles = 10.4 moles of Mg
Now that we know the number of moles of magnesium, we can calculate the mass of magnesium in grams using its molar mass. The molar mass of magnesium is 24.31 g/mol.
mass of Mg = moles of Mg × molar mass of Mg
mass of Mg = 10.4 moles × 24.31 g/mol = 252.98 grams of Mg
Therefore, 252.98 grams of magnesium metal will react completely with 5.2 liters of 4.0 M HCl.