how many grams of HCI are needed to make 0.3L of 4M solution

Question

need extreme help for my chemistry homework

Answer 1

To find how many grams of HCl to produce 0.3 L of a 4 M HCl solution, we need to use stoichiometry.

To find the grams of HCl needed, we start with the given, a 4 Molar solution of HCl, and only 0.3 Liters of it.

A concentration in M (Molarity) times the volume in Liters will give us moles, by this conversion factor:

(M)(L) [(1 mole / L ) / (1 M/1)] = [(M)(L)(mole)]/[(L)(M)] = moles

We also need to know the molar mass of HCl, which is the mass of one mole of HCl.

Here is my method for calculating the molar mass of any compound:

(1)(H) = (1)(1.00794 g) = 1.00794 g

(1)(Cl) = (1)(35.453 g) = +35.453 g

36.46094 g

Now we set up the stoichiometric problem:

x g HCl = (4 M HCl) (0.3 L) [(1 mole / L)/(1 M/1)] [(36.46094 g HCl)/(1 mole HCl)] = 43.75312 g HCl

43.75312 g of HCl is the exact answer, but not the significant figure answer.

The mass for Hydrogen has numbers with 5 places to the right of the decimal point, but the mass of chlorine only has numbers with 3 places to the right of the decimal point; so for this addition problem we could only use 36.461 g for the molar mass of HCl, which in itself is 5 significant figures.

We never round off any numbers until the very end of the problem, so we keep them all for now

There is only one significant figure in 4 M, and only one significant figure in 0.3 L.

The significant figure answer for this multiplication problem can be no more than the least number of significant figures in its factors, which is only one significant figure, so the correct significant figure answer for this problem is 40 grams of HCl.

Answer 2

Molarity = #mol solute / Liter of solution

The unit of the molarity is mol/L.

4 mol/L × 0.3 L = 1.2 mol HCl

Molar mass of HCl = 1 + 35.45 = 36.45 g/mol

The mass of HCl = 1.2 mol × 36.45 g/mol = 43.74 g HCl