Calculating Mass of Magnesium Metal

How many grams of magnesium metal will react completely with 6.3 liters of 4.5 M HCl?

What is the step-by-step process to determine the mass of magnesium metal required for the reaction?

Answer:

The mass of magnesium metal needed to completely react with 6.3 liters of 4.5 M HCl can be calculated by following these steps:

1. Determine the number of moles of HCl in 6.3 liters of 4.5 M solution:

Number of moles of HCl = 6.3 L * 4.5 mol/L = 28.35 mol

2. Use the balanced chemical equation to find the ratio between moles of HCl and moles of Mg:

2 HCl (aq) + Mg (s) → MgCl2 (aq) + H2 (g)

For every 2 moles of HCl, 1 mole of Mg is required. Set up a ratio:

2 mol HCl : 1 mol Mg

28.35 mol HCl : x mol Mg

Solve for x:

x = 28.35 mol HCl * 1 mol Mg / 2 mol HCl = 14.175 mol Mg

3. Convert moles of Mg to mass using the atomic mass of Mg:

14.175 mol Mg * 24.3 g Mg / mol Mg = 344.45 g

Therefore, 344.45 grams of magnesium metal will react completely with 6.3 liters of 4.5 M HCl.