How many Milliliters of 3.50 M HCl (aq)are required to react 5.75g of Zn(s)?
Zn + 2HCl ==> ZnCl2 + H2
mols Zn = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols Zn to mols HCl.
Then M HCl = mols HCl/L HCl. You know M and mols, solve for L HCl and convert to mL.
To find out how many milliliters of 3.50 M HCl (hydrochloric acid) are required to react with 5.75g of Zn (zinc), we need to use stoichiometry and the balanced equation of the reaction.
First, let's write the balanced chemical equation for the reaction between hydrochloric acid and zinc:
Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
This equation shows that one mole of zinc reacts with two moles of hydrochloric acid.
Next, we need to determine the moles of zinc present in 5.75g of Zn. We can use the molar mass of zinc to convert grams to moles:
Molar mass of Zn = 65.38 g/mol
Moles of Zn = mass (g) / molar mass (g/mol)
= 5.75 g / 65.38 g/mol
= 0.088 mol
Since two moles of HCl react with one mole of Zn, we need twice the number of moles of HCl as compared to Zn.
Moles of HCl = 2 * moles of Zn
= 2 * 0.088 mol
= 0.176 mol
Finally, we can use the molarity (M) and the definition of molarity to find the volume of 3.50 M HCl required:
Molarity (M) = moles / volume (L)
Volume (L) = moles / Molarity (M)
= 0.176 mol / 3.50 mol/L
= 0.05 L
Since the question asks for the volume in milliliters, we need to convert liters to milliliters:
Volume (mL) = Volume (L) * 1000
= 0.05 L * 1000
= 50 mL
Therefore, 50 milliliters of 3.50 M HCl (aq) are required to react with 5.75g of Zn(s).
To determine how many milliliters of 3.50 M HCl(aq) are required to react with 5.75g of Zn(s), we need to follow a few steps:
1. Write and balance the chemical equation for the reaction.
The balanced chemical equation for the reaction between hydrochloric acid (HCl) and zinc (Zn) is:
Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
2. Convert the mass of Zn(s) to moles.
To find the number of moles of zinc present in 5.75g, we divide the given mass by the molar mass of zinc (approximately 65.38 g/mol).
Moles of Zn = Mass of Zn / Molar mass of Zn
= 5.75g / 65.38 g/mol
3. Determine the stoichiometry of the reaction.
From the balanced equation, we can see that 1 mole of zinc reacts with 2 moles of hydrochloric acid.
4. Convert moles of Zn to moles of HCl.
Since the mole ratio between Zn and HCl is 1:2, we multiply the number of moles of Zn by 2.
Moles of HCl = Moles of Zn × 2
5. Calculate the volume of HCl solution.
The molarity (M) of the HCl solution is given as 3.50 M, which means there are 3.50 moles of HCl per liter (1000 mL) of solution.
To calculate the volume of HCl solution required, divide the moles of HCl by the molarity.
Volume of HCl solution = Moles of HCl / Molarity of HCl
Follow these steps to find the answer:
1. Calculate the number of moles of zinc using the given mass and molar mass of zinc.
2. Determine the number of moles of hydrochloric acid based on the stoichiometry of the reaction.
3. Convert moles of HCl to milliliters using the given molarity of the HCl solution.
Please note that this is a theoretical calculation, and in practice, you might have to consider factors like reaction efficiency and any limiting reagents.