Not quite it.
mols Zn = grams/molar mass (Zn, not ZnCl2)
Convert mols Zn to mols HCl. That 2 is right.
Then M HCl = mols HCl/L HCl. You know mols and M, solve for L ad convert to mL.
7.4/100.84g*2 mol
mols Zn = grams/molar mass (Zn, not ZnCl2)
Convert mols Zn to mols HCl. That 2 is right.
Then M HCl = mols HCl/L HCl. You know mols and M, solve for L ad convert to mL.
The molar mass of zinc is 65.38 g/mol. Thus, 7.4 g of zinc is equal to:
7.4 g / 65.38 g/mol = 0.113 mol
From the balanced chemical equation:
Zn + 2 HCl -> ZnCl2 + H2
It can be seen that 1 mole of zinc reacts with 2 moles of hydrochloric acid (HCl).
Therefore, the number of moles of HCl required to react with the given amount of zinc is:
0.113 mol Zn * 2 mol HCl / 1 mol Zn = 0.226 mol HCl
Now, you can calculate the volume of the 0.48 M HCl solution required to provide 0.226 mol of HCl using the formula:
Volume (in L) = moles / molarity
Volume (in L) = 0.226 mol HCl / 0.48 M HCl = 0.471 L
Finally, convert the volume from liters to milliliters by multiplying by 1000:
Volume (in mL) = 0.471 L * 1000 mL/L = 471 mL
Therefore, 471 mL of a 0.48 M HCl solution is needed to completely react with 7.4 g of zinc to form zinc(II) chloride.