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.

Zn + 2HCl ==> H2 + ZnCl2

mols Zn = grams Zn/atomic mass Zn = ?
Using the coefficients in the balanced equation, convert mols Zn to mols HCl. That will be mols Zn x 2 = mols HCl.

Then M = mols/L. You know M of HCl, mols HCl, solve for L HCl and convert to mL

To calculate the volume of the 0.48 M HCl solution needed to react completely with 7.4 g of zinc, we need to use the balanced chemical equation for the reaction:

Zn + 2HCl -> ZnCl2 + H2

From the balanced equation, we can see that 1 mole of zinc reacts with 2 moles of HCl to produce 1 mole of zinc(II) chloride.

1 mole of zinc is equal to its molar mass, which is 65.38 g.

To find the number of moles of zinc used, we can use the equation:

moles = mass / molar mass

moles of zinc = 7.4 g / 65.38 g/mol = 0.113 moles

Since the stoichiometry of the reaction is 1:2 (Zn:HCl), we need twice as many moles of HCl as zinc. Therefore, we need 2 * 0.113 = 0.226 moles of HCl.

Now, we can use the equation to find the volume of the HCl solution:

volume = moles / concentration

volume = 0.226 moles / 0.48 mol/L = 0.4708 L

To convert this to milliliters, we can multiply by 1000:

volume = 0.4708 L * 1000 mL/L = 470.8 mL

Therefore, approximately 470.8 milliliters of the 0.48 M HCl solution are needed to react completely with 7.4 g of zinc to form zinc(II) chloride.

To determine the volume of a solution needed to react completely with a given amount of a substance, you can use the concept of stoichiometry. In this case, we need to calculate the volume of a 0.48 M (Molar) HCl solution required to react with 7.4 g of zinc.

First, you need to convert the mass of zinc to moles. To do this, divide the given mass (7.4 g) by the molar mass of zinc (65.38 g/mol):

moles of zinc = 7.4 g / 65.38 g/mol = 0.113 moles

Since the balanced equation between zinc and hydrochloric acid is:
Zn + 2HCl -> ZnCl2 + H2

We can see that 1 mole of zinc reacts with 2 moles of hydrochloric acid to form 1 mole of zinc chloride. Therefore, the moles of hydrochloric acid needed will be twice the moles of zinc consumed:

moles of HCl = 2 * moles of zinc = 2 * 0.113 moles = 0.226 moles

Now we can use the molarity (M) of the hydrochloric acid solution to find the volume required to contain the calculated moles of HCl. The molarity is given as 0.48 M, which means there are 0.48 moles of HCl in 1 liter (1000 mL) of the solution.

To calculate the volume needed, we can use the equation:

moles = Molarity * Volume (in liters)

Rearranging the equation, we can find the volume:

Volume (in liters) = moles / Molarity

Substituting the values we have:

Volume (in liters) = 0.226 moles / 0.48 M = 0.4718 liters

Finally, to convert the volume from liters to milliliters, multiply by 1000:

Volume (in mL) = 0.4718 liters * 1000 mL/liter = 471.8 mL

Therefore, approximately 471.8 milliliters of a 0.48 M HCl solution are needed to react completely with 7.4 g of zinc to form zinc(II) chloride.