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.
csol = 4.20 j/g °c dsol = 1.02 g/ml
masssol = 102 g qrxn = -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: $25.0\ mL = 0.025\ L$.
Moles of HCl: $n_{HCl} = 1.00\ M \times 0.025\ L = 0.025\ mol$.

Step2: Determine moles of NaOH

Volume of NaOH: $75.0\ mL = 0.075\ L$.
Moles of NaOH: $n_{NaOH} = 1.00\ M \times 0.075\ L = 0.075\ mol$.

Step3: Identify limiting reactant

Reaction: $\ce{HCl + NaOH -> NaCl + H2O}$.
Mole ratio 1:1. HCl (0.025 mol) is limiting, so moles reacted = moles of HCl.

Answer:

0.025