if 0.10 L of 2.5 M NaOH react how many particles of HCI also react

Answer 1

moles = Molarity x Vol

mole NaOH = 2.5 M NaOH x 0.1 L = 0.25 moles NaOH

HCl + NaOH ----> NaCl + H₂O

Reacting Ration = 1:1, so moles of HCl reacted = 0.25 moles also.

1 mole = 6.022 x 10 ²³ number of particles

0.25 moles HCl will contain how many particles?

(0.25 moles x 6.022 x 10 ²³ number of particle)/1 mole

=1.5x10²³ number of particles of HCl also reacts!

Answer 2

I'm assuming this is the neutralization reaction between NaOH and HCl, which in this case, the balanced chemical equation is:

NaOH + HCl → NaCl + H₂O

Now, we perform the stoichiometry using our balanced chemical equation:

0.10 L x (2.5 mol NaOH/L) x (1 mol HCl/1 mol NaOH) x (6.022 x 10²³ HCl molecules/mol HCl) =

1.51 x 10²³ HCl molecules