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.
HCl = 1 + 35.5 = 36.5 grams/mol
so we need
22 grams /36.5 = .603 mol of HCl
.603 mol * 1000 mL/5.2 mol
= 116 mL
To find the volume of 5.2 M HCl needed to provide 22 g of HCl, we need to use the formula:
moles = mass / molar mass
First, let's calculate the number of moles of HCl needed:
moles = 22 g / molar mass of HCl
The molar mass of HCl (hydrochloric acid) is approximately 36.46 g/mol.
moles = 22 g / 36.46 g/mol
Next, we can use the molar concentration of the HCl solution to find the volume of 5.2 M HCl needed:
moles = volume (in liters) × Molarity
Since we want the answer in milliliters, we'll convert the volume from liters to milliliters:
moles = volume (in milliliters) × Molarity / 1000
Now, let's rearrange the formula to solve for the volume:
volume (in milliliters) = (moles * 1000) / Molarity
Substituting the calculated moles and the given molar concentration (5.2 M) into the formula:
volume (in milliliters) = (moles * 1000) / 5.2
Now, let's calculate it:
volume (in milliliters) = (22 g / 36.46 g/mol * 1000) / 5.2
volume (in milliliters) ≈ 645.077 mL
Therefore, approximately 645.077 mL of 5.2 M HCl must be transferred from the reagent bottle to provide 22 g of HCl for the reaction.
To find the number of milliliters of 5.2 M HCl needed, we can use the equation:
moles of solute = concentration × volume
First, we need to calculate the number of moles of HCl required for the reaction. We can do this using the formula:
moles = mass / molar mass
The molar mass of HCl is 36.5 g/mol. Therefore, the number of moles of HCl required for the reaction can be calculated as:
moles = 22 g / 36.5 g/mol
moles = 0.602 moles
Now, we can determine the volume of the 5.2 M HCl solution needed by rearranging the initial equation:
volume = moles of solute / concentration
volume = 0.602 moles / 5.2 M
volume ≈ 0.116 mL
Therefore, approximately 0.116 mL of the 5.2 M HCl solution must be transferred from the reagent bottle to provide 22 g HCl for the reaction.