Zinc reacts with hydrochloric acid according to the reaction equation:
Zn(s)+2HCl(aq)->ZnCl2(aq)+H2(g)
How many milliliters of 4.50M HCl1(aq) are required to react with 8.25g of an ore containing 30.0% Zn(s) by mass?
convert .30*8.25g to moles Zn
You need twice that number of moles of HCL
molesneeded=litersacid*4.5
conver that to ml.
5849
To find the number of milliliters of 4.50M HCl(aq) required to react with 8.25g of an ore containing 30.0% Zn(s) by mass, we need to follow these steps:
Step 1: Calculate the molar mass of Zn
The molar mass of Zn is 65.38 g/mol (you can find this information on the periodic table).
Step 2: Calculate the mass of Zn in the ore
To find the mass of Zn in the ore, multiply the total mass of the ore (8.25g) by the mass percentage of Zn (30.0% or 0.30):
Mass of Zn = 8.25g * 0.30 = 2.475g
Step 3: Convert the mass of Zn to moles
To convert the mass of Zn to moles, divide the mass of Zn by its molar mass:
Moles of Zn = Mass of Zn / Molar Mass of Zn = 2.475g / 65.38 g/mol = 0.03782 mol
Step 4: Determine the stoichiometry of the reaction
From the balanced equation, we can see that the stoichiometric ratio between Zn and HCl is 1:2. This means that it takes 2 moles of HCl to react with 1 mole of Zn.
Step 5: Calculate the number of moles of HCl required
Since the stoichiometric ratio is 1:2, the number of moles of HCl required is twice the number of moles of Zn:
Moles of HCl = 2 * Moles of Zn = 2 * 0.03782 mol = 0.07564 mol
Step 6: Calculate the volume of HCl solution
To calculate the volume, we need to use the molarity of the HCl solution. The equation relating moles, volume, and molarity is:
Moles = Molarity * Volume
Rearranging the equation gives:
Volume = Moles / Molarity
So, to find the volume of HCl solution required, divide the number of moles of HCl (0.07564 mol) by the molarity of HCl (4.50 mol/L):
Volume of HCl solution = Moles of HCl / Molarity of HCl = 0.07564 mol / 4.50 mol/L = 0.01681 L
Step 7: Convert the volume to milliliters
Since the volume was calculated in liters, convert it to milliliters by multiplying by 1000:
Volume of HCl solution in milliliters = Volume of HCl solution in liters * 1000 = 0.01681 L * 1000 = 16.81 mL
Therefore, approximately 16.81 milliliters of 4.50M HCl(aq) are required to react with 8.25g of the ore containing 30.0% Zn(s) by mass.