Answer: 27.1 milliliters of 2.00 M HCl(aq) are required to react with 4.55 g of the ore containing 39.0% Zn(s) by mass.
Step-by-step explanation:
To determine how many milliliters of 2.00 M HCl(aq) are required to react with 4.55 g of an ore containing 39.0% Zn(s) by mass, we need to use stoichiometry and the concept of molar ratios.
1. Find the molar mass of Zn:
The molar mass of Zn is 65.38 g/mol.
2. Calculate the mass of Zn in the ore:
Given that the ore contains 39.0% Zn by mass, we can calculate the mass of Zn in 4.55 g of the ore:
Mass of Zn = 4.55 g * 0.39 = 1.77345 g Zn
3. Convert the mass of Zn to moles:
Using the molar mass of Zn, we can convert the mass of Zn to moles:
Moles of Zn = 1.77345 g Zn / 65.38 g/mol = 0.0271 mol Zn
4. Use the balanced chemical equation:
The balanced chemical equation for the reaction between HCl and Zn is:
Zn + 2 HCl → ZnCl2 + H2
From the balanced equation, we can see that the ratio between Zn and HCl is 1:2.
5. Determine the moles of HCl required:
Since the molar ratio between Zn and HCl is 1:2, the moles of HCl required will be twice the moles of Zn:
Moles of HCl = 2 * 0.0271 mol Zn = 0.0542 mol HCl
6. Calculate the volume of HCl using the molarity:
We are given that the HCl solution is 2.00 M. Molarity is defined as moles of solute per liter of solution. Rearranging the equation, we can solve for the volume:
Volume of HCl = Moles of HCl / Molarity = 0.0542 mol HCl / 2.00 mol/L = 0.0271 L HCl
7. Convert liters to milliliters:
1 L = 1000 mL
Volume of HCl = 0.0271 L * 1000 mL/L = 27.1 mL HCl
Therefore, 27.1 milliliters of 2.00 M HCl(aq) are required to react with 4.55 g of the ore containing 39.0% Zn(s) by mass.
I hope this helps :)