What volume of 0.140 M HCl is needed to neutralize 2.93 g of Mg(OH)2
Well, this sounds like a job for Chemistry Man! But unfortunately, he called in sick today, so you're stuck with me, Clown Bot.
Let's put on our thinking caps and figure this out. First, we need to calculate the moles of Mg(OH)2 that we have.
The molar mass of Mg(OH)2 is calculated by adding up the atomic masses of magnesium (24.31 g/mol), oxygen (16.00 g/mol), and hydrogen (1.01 g/mol) - giving us a total of 58.33 g/mol.
Next, we can use the equation:
moles = mass / molar mass
Plugging in the values we have, we get:
moles = 2.93 g / 58.33 g/mol
Calculating this, we find that we have approximately 0.05 moles of Mg(OH)2.
The balanced equation for the reaction between Mg(OH)2 and HCl is:
Mg(OH)2 + 2HCl -> MgCl2 + 2H2O
From the equation, we can see that 1 mole of Mg(OH)2 reacts with 2 moles of HCl. Therefore, we need twice as many moles of HCl as we have of Mg(OH)2.
Since we have 0.05 moles of Mg(OH)2, we need 0.1 moles of HCl.
Now, we can calculate the volume of 0.140 M HCl needed.
The equation for molarity is:
moles = Molarity x Volume (in liters)
We have moles (0.1 moles) and the molarity of the HCl solution (0.140 M), so we can rearrange the equation to find the volume:
Volume (in liters) = moles / Molarity
Plugging in the values, we get:
Volume (in liters) = 0.1 moles / 0.140 M
Calculating this, we find that approximately 0.71 liters (or 710 milliliters) of 0.140 M HCl is needed to neutralize 2.93 g of Mg(OH)2.
I hope that was helpful! Now remember, never invite a clown to a chemistry party because they'll always throw in a silly solution!
To determine the volume of 0.140 M HCl needed to neutralize 2.93 g of Mg(OH)2, we can use the concept of stoichiometry. The balanced chemical equation for the reaction between HCl and Mg(OH)2 is as follows:
2 HCl + Mg(OH)2 → MgCl2 + 2 H2O
From the equation, we can see that it takes two moles of HCl to react with one mole of Mg(OH)2. Therefore, we need to convert the mass of Mg(OH)2 into moles first.
Step 1: Calculate the molar mass of Mg(OH)2:
Mg: atomic mass = 24.31 g/mol
O: atomic mass = 16.00 g/mol
H: atomic mass = 1.01 g/mol
Molar mass of Mg(OH)2 = (24.31 g/mol) + 2[(1.01 g/mol) + (16.00 g/mol)] = 58.33 g/mol
Step 2: Convert the mass of Mg(OH)2 into moles:
Moles of Mg(OH)2 = (2.93 g) / (58.33 g/mol) = 0.0502 mol
Step 3: Determine the volume of 0.140 M HCl needed based on stoichiometry:
From the balanced equation, we know that 2 moles of HCl react with 1 mole of Mg(OH)2.
Therefore, moles of HCl = 2 × (0.0502 mol) = 0.1004 mol
Now we can use the definition of molarity to find the volume of HCl needed:
Molarity (M) = moles / volume (L)
Rearranging the equation, we have:
Volume (L) = moles / molarity (M) = 0.1004 mol / 0.140 M = 0.717 L
Finally, we convert the volume from liters to milliliters:
Volume (mL) = 0.717 L × (1000 mL / 1 L) = 717 mL
Therefore, approximately 717 mL of the 0.140 M HCl solution is needed to neutralize 2.93 g of Mg(OH)2.
To find the volume of 0.140 M HCl needed to neutralize 2.93 g of Mg(OH)2, we first need to determine the number of moles of Mg(OH)2.
Step 1: Calculate the molar mass of Mg(OH)2.
Molar mass of Mg(OH)2 = atomic mass of Mg + (2 * atomic mass of O) + (2 * atomic mass of H)
= 24.31 g/mol + (2 * 16.00 g/mol) + (2 * 1.01 g/mol)
= 58.33 g/mol
Step 2: Convert the mass of Mg(OH)2 to moles.
Number of moles = mass / molar mass
= 2.93 g / 58.33 g/mol
≈ 0.0502 mol
Step 3: Write the balanced chemical equation for the reaction between Mg(OH)2 and HCl.
Mg(OH)2 + 2HCl -> MgCl2 + 2H2O
Step 4: Use the stoichiometry of the balanced equation to determine the mole ratio between Mg(OH)2 and HCl.
From the balanced equation, we can see that 1 mole of Mg(OH)2 reacts with 2 moles of HCl.
Step 5: Calculate the number of moles of HCl needed.
Number of moles of HCl = 2 * number of moles of Mg(OH)2
= 2 * 0.0502 mol
= 0.1004 mol
Step 6: Now, we need to find the volume of 0.140 M HCl corresponding to 0.1004 moles. The formula for finding volume (V) is:
V = moles / concentration
V = 0.1004 mol / 0.140 mol/L
≈ 0.717 L
Therefore, approximately 0.717 liters (or 717 mL) of 0.140 M HCl is needed to neutralize 2.93 g of Mg(OH)2.
Mg(OH)2 + 2 HCl ==> MgCl2 + 2H2O
mols Mg(OH)2 = grams/molar mass = ?
mols HCl = twice that (from the coefficients in the balanced equation.
Then molarity HCl = mols HCl/L HCl.. You know molarity and mols, solve for L. Convert to mL if needed.