A 0.042 mol sample of Ca(OH)2 requires 51.21 mL of aqueous HCl for neutralization. What is the concentration of the HCl

Conc = 1.64 Mol / dm3

Explanation:

We have to first put down the equation of the neutralization reaction. This is given as;

2HCl + Ca(OH)2 --> CaCl2  + 2H2O

This means that 2 mol of HCl requires 1 mol of Ca(OH)2.

2 = 1

x = 0.042

x = 0.084 mol of HCl required.

Volume = 51.21 ml = 0.05121 dm3

Concentration = Mass / Volume

Conc = 0.084 / 0.05121

Conc = 1.64 Mol / dm3

the answer is true.

sorry i was late, but i don't know what assignment your talking about, i'm in flvs as well, im in earth space science. but u are behind pace bc every class should be in the 3's or 4'