Answer:
For this problem we just need to remember the equation and that the volume is always in liters: MaVa=MbVb
Ma= 1.338 mol/L Va= 18.75 mL= 0.01875 L Mb= x Vb=24.73 mL= 0.02473 L
So now we can plug into the equation and solve:
1.338 mol/L * 0.01875 L= x mol/L * 0.02473 L
This is a two step process: stoichiometry and using the answer from the first part to plug and chug it into the Molariy equation.
First step: setup the stoichiometry problem:
3.1171 g Na2CO3 * (1 mol/106 g) * (2 mol HCl/ 1 mol Na2CO3)= mol HCl
Second step: Molarity equation
Molarity= moles HCl/ L M= mol HCl/0.04027 L
For the third problem, you will just use the same equation as the first: MaVa= MbVb
Ma=0.57 M Va= x L Mb= 0.875 M Vb= 23.83 mL= 0.02383 L
0.57 M * x L= 0.875 M * 0.02383 L
With this equation, we want to find the moles of NaOH by using the molarity equation first then because MaVa= MbVb, we know that the number of moles has to be equal.
0.75 M= x mol/ 0.0227 L mol NaOH = 0.75 M *0.0227 L mol NaOH= 0.017025
So next, we can set it to the mass of the acid using this equation: Molar Mass= mass/ moles
Molar Mass= 3.6 g/ 0.017025 mol
And with that you will find the molar mass of HX, and even determine what X is.
Explanation:
hope I helped
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