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) + 2 HCl (aq) yields MgCl2 (aq) + H2 (g)
mols HCl = M x L = ?
Using the coefficients in the balanced equation, convert mols HCl to mols Mg.
Now convert mols Mg to g. g = mols x atomic mass.
To solve this problem, we need to use stoichiometry and the given information about the concentration of HCl.
Step 1: Write and balance the chemical equation:
Mg (s) + 2 HCl (aq) → MgCl2 (aq) + H2 (g)
Step 2: Determine the number of moles of HCl:
To find the number of moles, we can use the formula:
moles = concentration (M) × volume (L)
Given that the concentration is 4.0 M and the volume is 5.2 L:
moles of HCl = 4.0 M × 5.2 L = 20.8 moles
Step 3: Use stoichiometry to find the number of moles of Mg:
From the balanced equation, we can see that the mole ratio between Mg and HCl is 1:2. This means that for every mole of Mg, two moles of HCl are needed.
moles of Mg = (moles of HCl) ÷ 2
moles of Mg = 20.8 moles ÷ 2 = 10.4 moles
Step 4: Convert moles of Mg to grams:
To convert the moles of Mg to grams, we need to use its molar mass. The molar mass of Mg 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 = 252.424 g
Therefore, 252.424 grams of magnesium metal will react completely with 5.2 liters of 4.0 M HCl.