How many milliliters of a 0.81 M HCl solution are needed to react completely with 2.2 g of zinc to form zinc(II) chloride?
Write the equation and balance it.
Convert 2.2 g Zn to mols. mols = grams/molar mass.
Use the coefficients in the balanced equation to convert mols Zn to mols HCl.
M HCl = mols HCl/L HCl. You know moles and M, solve for L HCl and convert to mL.
To find how many milliliters of the 0.81 M HCl solution are needed to react completely with 2.2 g of zinc, we need to use stoichiometry and the balanced chemical equation.
The balanced chemical equation for the reaction between hydrochloric acid (HCl) and zinc (Zn) is:
Zn + 2HCl -> ZnCl2 + H2
From the balanced equation, we can see that for every 1 mole of Zn, 2 moles of HCl are required. Therefore, we need to convert the mass of zinc (2.2 g) into moles.
To calculate the number of moles of zinc, we need to divide the given mass by the molar mass of zinc (65.38 g/mol):
Number of moles of Zn = mass of Zn / molar mass of Zn
Number of moles of Zn = 2.2 g / 65.38 g/mol ≈ 0.0337 mol
Since the stoichiometric ratio is 1:2 (one mole of zinc reacts with two moles of HCl), we can calculate the number of moles of HCl required using the same ratio:
Number of moles of HCl = 2 * Number of moles of Zn
Number of moles of HCl = 2 * 0.0337 mol = 0.0674 mol
Now, we have the number of moles of HCl required for the reaction. To find the volume (in milliliters) of the 0.81 M HCl solution needed, we can use the molarity formula:
Molarity = Number of moles / Volume (in liters)
Rearranging the formula, we get:
Volume (in liters) = Number of moles / Molarity
Volume (in liters) = 0.0674 mol / 0.81 mol/L ≈ 0.0832 L
To convert from liters to milliliters, we multiply the volume by 1000:
Volume (in milliliters) = 0.0832 L * 1000 mL/L ≈ 83.2 mL
Therefore, approximately 83.2 milliliters of the 0.81 M HCl solution are needed to react completely with 2.2 g of zinc to form zinc(II) chloride.