How many milliliters of 5.5 M HCl must be
transfered from a reagent bottle to provide
23 g HCl for a reaction?
Answer in units of mL
Andy! STOP CHEATING!!!
To find the number of milliliters (mL) of 5.5 M HCl needed to provide 23 g of HCl, we can use the equation:
moles = mass / molar mass
First, we need to calculate the moles of HCl using the given mass of 23 g and the molar mass of HCl, which is 36.46 g/mol.
moles = 23 g / 36.46 g/mol
moles ≈ 0.631 mol
The molarity (M) of a solution is defined as moles of solute dissolved per liter of solution. So, to find the volume in liters, we divide moles by the molarity:
volume (in liters) = moles / Molarity
volume = 0.631 mol / 5.5 M
volume ≈ 0.115 L
Finally, to convert from liters to milliliters, we multiply the volume in liters by 1000:
volume (in mL) = 0.115 L * 1000 mL/L
volume ≈ 115 mL
Therefore, you would need approximately 115 mL of 5.5 M HCl from the reagent bottle to provide 23 g of HCl for the reaction.
mols x molar mass = grams.
You have grams HCl and molar mass HCl,
substitute and solve for mols HCl.
Then M x L = mols. You have mols and M, solve for L and convert to mL.