Zinc reacts with hydrochloric acid according to the reaction equation...

Zn (s) + 2HCl (aq) ---> ZnCl2 (aq) + H2 (g)

How many milliliters of 5.00 M HCl (aq) are required to react with 5.55 g of Zn (s)??

mols Zn = grams/molar mass = ?

Using the coefficients in the balanced equation, convert mols Zn to mols HCl.
Then M HCl = mols HCl/L HCl. You know mols and M, solve for L and convert to mL.

.849mL?

It's 33.9mL!!!!!!

:) Thanks

I obtained 33.96 which I would round to 34.0 to three significant figures. I suspect you just threw the 6 in 33.96 away without thinking about it.

To find the volume of 5.00 M hydrochloric acid (HCl) required to react with 5.55 g of zinc (Zn), you can follow these steps:

Step 1: Determine the molar mass of Zn
The molar mass of zinc (Zn) is 65.38 g/mol.

Step 2: Convert grams of Zn to moles
To convert grams of Zn to moles, divide the given mass by the molar mass.
moles of Zn = mass of Zn / molar mass of Zn
moles of Zn = 5.55 g / 65.38 g/mol = 0.0848 mol

Step 3: Use stoichiometry to find the moles of HCl required
From the balanced equation, we can see that the stoichiometric ratio between Zn and HCl is 1:2. This means that 1 mole of Zn reacts with 2 moles of HCl.
moles of HCl = 2 * moles of Zn
moles of HCl = 2 * 0.0848 mol = 0.1696 mol

Step 4: Use the definition of molarity to find the volume of HCl
The given concentration of HCl is 5.00 M, which means that there are 5.00 moles of HCl in 1 liter of solution.
volume of HCl (in liters) = moles of HCl / molarity of HCl
volume of HCl (in liters) = 0.1696 mol / 5.00 mol/L = 0.0339 L

Step 5: Convert the volume of HCl to milliliters
To convert liters to milliliters, multiply the volume in liters by 1000.
volume of HCl (in milliliters) = volume of HCl (in liters) * 1000
volume of HCl (in milliliters) = 0.0339 L * 1000 = 33.9 mL

Therefore, approximately 33.9 milliliters (mL) of 5.00 M HCl are required to react with 5.55 grams (g) of Zn.