The UNBALANCED equation is: Zn + HCl –] ZnCl2 + H2 How many milliliters of a 0.48 M HCl solution are needed to react completely with 7.4 g of zinc to form zinc(II) chloride? Answer in units of mL.
7.4/100.84g*2 mol
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.
To find the number of moles of zinc (Zn) present in 7.4 g, you need to divide the given mass by the molar mass of zinc.
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.