How many grams of MgCl2 are produced when 650 mL of 3 M hydrochloric acid solution reacted with excess magnesium hydroxide solution, Mg(OH)2?
Mg(OH)2 + 2HCl = MgCl2 + 2H2O
0.650L * 3mol/L = 1.95 moles HCl
you get half that many moles of MgCl2
so how many grams of MgCl2 in 0.975 moles?
To determine the amount of magnesium chloride (MgCl2) produced, we need to first balance the chemical equation for the reaction between hydrochloric acid (HCl) and magnesium hydroxide (Mg(OH)2):
2HCl + Mg(OH)2 -> MgCl2 + 2H2O
From the balanced equation, we can see that 2 moles of hydrochloric acid react with 1 mole of magnesium hydroxide to produce 1 mole of magnesium chloride and 2 moles of water.
Step 1: Calculate the number of moles of hydrochloric acid (HCl) in the 650 mL of 3 M solution.
- Moles of HCl = (Volume of solution in liters) x (Concentration of HCl)
- Convert the volume from milliliters to liters:
Volume of solution = 650 mL = 650/1000 = 0.65 L
- Calculate the moles of HCl:
Moles of HCl = 0.65 L x 3 M = 1.95 moles
Step 2: Determine the number of moles of magnesium hydroxide (Mg(OH)2) required for the reaction.
- According to the balanced equation, 2 moles of HCl react with 1 mole of Mg(OH)2.
- Therefore, the moles of Mg(OH)2 = (Moles of HCl) / 2 = 1.95 moles / 2 = 0.975 moles
Step 3: Calculate the molar mass of magnesium chloride (MgCl2).
- The molar mass of Mg is 24.31 g/mol, and the molar mass of Cl is 35.45 g/mol.
- The molar mass of MgCl2 = (24.31 g/mol) + (2 x 35.45 g/mol) = 95.21 g/mol
Step 4: Determine the mass of magnesium chloride (MgCl2) produced.
- The mass of MgCl2 = (Moles of MgCl2) x (Molar mass of MgCl2)
- Mass of MgCl2 = 0.975 moles x 95.21 g/mol = 92.71 grams
Therefore, approximately 92.71 grams of MgCl2 are produced when 650 mL of 3 M hydrochloric acid solution reacts with excess magnesium hydroxide solution, Mg(OH)2.
To solve this problem, you need to use the concept of stoichiometry and the balanced chemical equation of the reaction between hydrochloric acid (HCl) and magnesium hydroxide (Mg(OH)2).
The balanced chemical equation for the reaction is:
2 HCl + Mg(OH)2 → MgCl2 + 2 H2O
According to the balanced equation, 2 moles of HCl react with 1 mole of Mg(OH)2 to produce 1 mole of MgCl2.
Here's how you can calculate the number of moles of MgCl2 produced:
Step 1: Convert the volume of the HCl solution from milliliters (mL) to liters (L):
650 mL × (1 L / 1000 mL) = 0.65 L
Step 2: Use the given concentration of the HCl solution (3 M) to calculate the number of moles of HCl:
0.65 L × 3 mol/L = 1.95 moles of HCl
Step 3: According to the balanced equation, 2 moles of HCl react with 1 mole of Mg(OH)2 to produce 1 mole of MgCl2. Therefore, the moles of MgCl2 produced will be half of the moles of HCl used:
1.95 moles of HCl / 2 = 0.975 moles of MgCl2
Step 4: Finally, calculate the mass of MgCl2 produced using its molar mass:
Molar mass of MgCl2 = (24.31 g/mol for Mg) + (2 * 35.45 g/mol for Cl) = 95.21 g/mol
Mass of MgCl2 produced = 0.975 moles × 95.21 g/mol = 92.77 grams
Therefore, approximately 92.77 grams of MgCl2 are produced when 650 mL of 3 M hydrochloric acid solution reacts with excess magnesium hydroxide solution.