Year 11 Moles (1 Viewer)

[Gtsh]

How do you this question? I can't seem to find g/100ml.

25.0mL of a solution of calcium hydroxide required 8.13mL of a 0.102mol L −1 solution of hydrochloric acid solution for reaction. Calculate the concentration of calcium hydroxide in mol and in g/100mL. The products of the reaction are calcium chloride solution and water.

 

[Edzion education]

I have put resources for this style of question on my website in the past, on my website this section is in year 12 chemistry, titration questions.

Firstly,
Ca(OH)2 + 2HCl --> CaCl2 + 2H2O

Since, 8.13mL of a 0.102mol L −1 of HCl was used, we can find the number of moles of HCl using n = M*v
0.00813 L * 0.102mol L −1 = 0.00082926 mol of HCl

From our above chemical equation, this means for every 2 moles of HCl, we will have had 1 mol of Ca(OH)2, so 0.00082926 mol / 2 will equal the 0.00041463 moles of Ca(OH)2
4.1463 * 10^(-4) moles of Ca(OH)2

The concentration of the Ca(OH)2 present is then going to be, 4.1463 * 10^(-4) moles of Ca(OH)2 / 0.025L = 0.0165852 mol/L

Finally to find the g/ L, multiply by the molar mass of Ca(OH)2, which is 74.09 g/mol * 0.0165852 mol/L = 1.228797468 g/1000 mL

and divide by 10 to make it g/ 100mL

0.123 g/100 mL of Ca(OH)2

Hope this helps!
Edzion education

 

[Gtsh]

That makes a lot of sense. Thank you for your help!