Zinc metal reacts with hydrochloric acid to produce zinc (II) chloride and hydrogen gas. what volume (in mL) of 2.0M HCl is required to dissolve a 10.0g piece of zinc?
To find the volume of hydrochloric acid required, we need to use the concept of stoichiometry.
Step 1: Write a balanced equation for the reaction:
Zn (s) + 2HCl (aq) -> ZnCl2 (aq) + H2 (g)
According to the equation, one mole of zinc reacts with 2 moles of hydrochloric acid to form one mole of zinc chloride and one mole of hydrogen gas.
Step 2: Convert the mass of zinc to moles using its molar mass:
Molar mass of Zn = 65.38 g/mol
Number of moles of Zn = mass (g) / molar mass (g/mol)
Number of moles of Zn = 10.0 g / 65.38 g/mol = 0.153 mol
Step 3: Use the stoichiometry of the balanced equation to determine the moles of HCl needed:
From the balanced equation, we see that 1 mole of Zn reacts with 2 moles of HCl.
Therefore, moles of HCl = 2 * moles of Zn = 2 * 0.153 mol = 0.306 mol
Step 4: Convert moles of HCl to volume in liters using the molarity:
Molarity = moles of solute / volume of solution (in liters)
0.306 mol / 2.0 mol/L = volume (in L)
Step 5: Convert the volume from liters to milliliters:
1 L = 1000 mL
So, volume (in mL) = volume (in L) * 1000
volume (in mL) = (0.306 mol / 2.0 mol/L) * 1000 = 153 mL
Therefore, 153 mL of 2.0M HCl is required to dissolve a 10.0g piece of zinc.