What volume of 3M hydrochloric acid will be required to react completely with 3.25 g of zinc
Now
impatient much?
how many moles of Zn in 3.25g?
The equation will tell you how many moles of HCl will be used
3M HCl = 3 moles/liter, so divide the moles HCl by 3 to get liters
Zn + 2HCl ==> ZnCl2 + H2
mols Zn = grams/atomic mass = 3.25/65.4 = about 0.0496 but you need to confirm the atomic mass Zn and give a more accurate answer.
Then from the equation, that many mols Zn will require twice that many mols of HCl.
Finally, mols = M HCl x L HCl. You know mols HCl and M HCl, solve for L HCl. Convert to mL if desired.
Post your work if you get stuck.
To determine the volume of 3M hydrochloric acid required to react completely with 3.25 g of zinc, we need to use stoichiometry and the balanced chemical equation for the reaction.
The balanced equation for the reaction between zinc (Zn) and hydrochloric acid (HCl) is:
Zn + 2HCl → ZnCl2 + H2
According to the balanced equation, 1 mole of zinc reacts with 2 moles of hydrochloric acid to produce 1 mole of zinc chloride and 1 mole of hydrogen gas.
Step 1: Convert the mass of zinc to moles.
The molar mass of zinc is 65.38 g/mol. Divide the given mass of zinc by its molar mass to obtain the number of moles:
3.25 g zinc × (1 mol/65.38 g) = 0.0498 mol zinc
Step 2: Determine the moles of hydrochloric acid needed.
Since the stoichiometric ratio between zinc and hydrochloric acid is 1:2, the number of moles of hydrochloric acid required will be twice the number of moles of zinc.
Number of moles of HCl = 2 × 0.0498 mol zinc = 0.0996 mol HCl
Step 3: Calculate the volume of hydrochloric acid using the molarity.
The molarity (M) of hydrochloric acid is given as 3M, which means there are 3 moles of HCl in 1 liter of solution.
Volume of HCl (in liters) = number of moles of HCl / molarity
Volume of HCl = 0.0996 mol / 3 M = 0.0332 L = 33.2 mL
Therefore, you will need approximately 33.2 mL of 3M hydrochloric acid to react completely with 3.25 g of zinc.