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.