How many milliliters of 5.2 M HCl must be

transfered from a reagent bottle to provide
22 g HCl for a reaction?
Answer in units of mL

How many moles do you need for 22 g? That is moles = grams/molar mass. Solve for moles.

Then M HCl = moles HCl/L HCl.
Solve for L and convert to mL.

To find the number of milliliters of 5.2 M HCl required to provide 22 g of HCl, we need to use the equation:

moles = mass / molar mass

First, we need to find the number of moles of HCl:

moles = 22 g / molar mass of HCl

The molar mass of HCl is calculated as follows:

Molar mass of HCl = 1(atomic mass of hydrogen) + 1(atomic mass of chlorine)

The atomic masses of hydrogen (H) and chlorine (Cl) are approximately 1 g/mol and 35.5 g/mol, respectively.

Molar mass of HCl = 1(1 g/mol) + 1(35.5 g/mol) = 36.5 g/mol

Now we can calculate the number of moles of HCl:

moles = 22 g / 36.5 g/mol

Next, we will use the formula:

Molarity = moles / volume

Rearranging the equation, we have:

Volume = moles / Molarity

The molarity, in this case, is 5.2 M (given in the question). So, we can find the volume in milliliters:

Volume (in mL) = (moles / Molarity) * 1000 mL

Substituting the values, we get:

Volume (in mL) = (moles * 1000) / Molarity

Now we substitute the value of moles (calculated earlier) and the Molarity of 5.2 M into the equation:

Volume (in mL) = (moles * 1000) / 5.2 M

Plug in the calculated value for moles and solve for volume to get the answer in units of mL.

To find the volume of 5.2 M HCl needed, we can use the equation:

Molarity(M1) * Volume(V1) = Molarity(M2) * Volume(V2)

Given:
M1 = 5.2 M
M2 = concentration is not given, but we need to find V2
V1 = ?
V2 = 22 g

First, we need to convert the mass of HCl to moles using its molar mass. The molar mass of HCl is 36.461 g/mol.

Number of moles of HCl = mass/molar mass
= 22 g / 36.461 g/mol
= 0.603 moles

Now we can substitute the given values into the equation and solve for V1:

5.2 M * V1 = 0.603 moles

V1 = 0.603 moles / 5.2 M

V1 ≈ 0.116 mL

Therefore, approximately 0.116 mL of 5.2 M HCl must be transferred from the reagent bottle to provide 22 g of HCl for the reaction.