Question

hcl + naoh → nacl + h₂o
25.0 ml of 1.00 m hcl is added to
75.0 ml of 1.00 m naoh. the
resulting temperature change is an
increase of 8.65 °c.
$c_{sol} = 4.20 j/g °c$ $d_{sol} = 1.02 g/ml$
$mass_{sol} = 102 g$ $q_{rxn} = -3,710 j$
how many moles reacted?
moles reacted = ? mol

Explanation:

Step1: Determine moles of HCl

Molarity formula: \( M = \frac{n}{V} \), so \( n = M \times V \).
Volume of HCl: \( 25.0\space mL = 0.0250\space L \), \( M = 1.00\space M \).
\( n_{HCl} = 1.00\space mol/L \times 0.0250\space L = 0.0250\space mol \).

Step2: Determine moles of NaOH

Volume of NaOH: \( 75.0\space mL = 0.0750\space L \), \( M = 1.00\space M \).
\( n_{NaOH} = 1.00\space mol/L \times 0.0750\space L = 0.0750\space mol \).

Step3: Identify limiting reactant

Reaction: \( HCl + NaOH
ightarrow NaCl + H_2O \) (1:1 stoichiometry).
HCl (0.0250 mol) is less than NaOH (0.0750 mol), so HCl is limiting. Moles reacted = moles of limiting reactant.

Answer:

0.0250