How many grams MgCl2 will be formed from 25.6 mL of a 0.100M HCl solution reacting with excess magnesium metal
Ah, chemistry, the art of mixing potions! Now, let's see what magic we can conjure up here. First things first, we need to write a balanced equation to work our comedic chemistry magic:
2HCl + Mg -> MgCl2 + H2
Now, let's calculate the moles of HCl in 25.6 mL of a 0.100M solution. Using the equation Molarity (M) = moles/volume (L), we can convert mL to L:
25.6 mL = 0.0256 L
Moles of HCl = Molarity x Volume
= 0.100 mol/L x 0.0256 L
= 0.00256 mol
According to our balanced equation, 2 moles of HCl will produce 1 mole of MgCl2. So, if we have 0.00256 mol of HCl, we can expect half that amount, 0.00128 mol, of MgCl2 to form.
To find grams, yep, grams, of MgCl2, we need to use its molar mass. The molar mass of MgCl2 is approximately 95.211 grams/mole.
Mass (grams) = Moles x Molar mass
= 0.00128 mol x 95.211 g/mol
≈ 0.122 grams
So, approximately 0.122 grams of MgCl2 will be formed, or as I like to call it, a pinch of magnesium chloride! Voila!
To determine the number of grams of MgCl2 formed, we need to use the balanced chemical equation for the reaction between HCl and magnesium metal:
2 HCl + Mg → MgCl2 + H2
From the balanced equation, we can see that 2 moles of HCl react with 1 mole of magnesium to form 1 mole of MgCl2. We will follow these steps to calculate the mass of MgCl2 formed:
Step 1: Determine the moles of HCl used
- Volume of HCl solution = 25.6 mL = 0.0256 L
- Concentration of HCl solution = 0.100 M
- Moles of HCl = Volume × Concentration = 0.0256 L × 0.100 mol/L = 0.00256 mol
Step 2: Determine the moles of MgCl2 formed
- From the balanced equation, we know that 2 moles of HCl react with 1 mole of MgCl2.
- Therefore, the moles of MgCl2 formed will be the same as the moles of HCl used.
- Moles of MgCl2 = 0.00256 mol
Step 3: Calculate the mass of MgCl2 formed
- The molar mass of MgCl2 = atomic mass of Mg + 2 × atomic mass of Cl
- The atomic mass of Mg = 24.31 g/mol
- The atomic mass of Cl = 35.45 g/mol
- Molar mass of MgCl2 = (24.31 g/mol) + 2 × (35.45 g/mol) = 95.21 g/mol
- Mass of MgCl2 formed = Moles of MgCl2 × Molar mass of MgCl2
= 0.00256 mol × 95.21 g/mol = 0.244 g
Therefore, 0.244 grams of MgCl2 will be formed from 25.6 mL of a 0.100M HCl solution reacting with excess magnesium metal.
To determine the number of grams of MgCl2 formed, we need to use stoichiometry. By balancing the chemical equation for the reaction of HCl with magnesium metal, we have:
2 HCl + Mg -> MgCl2 + H2
From the balanced equation, we can see that the stoichiometric ratio between HCl and MgCl2 is 2:1. This means that for every 2 moles of HCl reacted, 1 mole of MgCl2 is formed.
First, let's convert the volume of the HCl solution to moles using the given concentration:
Molarity (M) = moles/volume (L)
0.100 M = moles/0.0256 L
moles = 0.100 M * 0.0256 L = 0.00256 moles of HCl
Since the stoichiometric ratio tells us that 2 moles of HCl react to form 1 mole of MgCl2, we can determine the moles of MgCl2 formed:
moles of MgCl2 = 0.00256 moles of HCl / 2 = 0.00128 moles of MgCl2
Finally, we can convert moles of MgCl2 to grams using its molar mass:
Molar mass of MgCl2 = 24.305 g/mol + 2 * 35.453 g/mol = 95.211 g/mol
grams of MgCl2 = moles of MgCl2 * molar mass of MgCl2
grams of MgCl2 = 0.00128 moles * 95.211 g/mol ≈ 0.122 grams
Therefore, approximately 0.122 grams of MgCl2 will be formed.
the solution contains
.0256L * .100mol/L = .00256 mol
Mg + 2HCl = MgCl2 + H2
each 2 moles of HCl produces 1 mole of MgCl2
So, you get .00128 moles of MgCl2
or, .00128 * 59.76 = 0.076g of MgCl2