Question

2naoh + h₂so₄ → 2h₂o + na₂so₄
50.0 ml of 0.50 m h₂so₄ is added to
60.0 ml of 0.80 m naoh. the resulting
temperature change is an increase of
1.20 °c.
$c_{sol}$ = 4.20 j/g °c $d_{sol}$ = 1.05 g/ml $c_{cal}$ = 4.50 j/°c
$mass_{sol}$ = 116 g $q_{rxn}$ = -590. j
how many moles of na₂so₄ will form?
moles$_{rxn}$ = ? mol
hint: what is the limiting reactant?

Explanation:

Step1: Calculate moles of \( H_2SO_4 \)

Molarity formula: \( n = M \times V \) (in L). \( V_{H_2SO_4} = 50.0 \, mL = 0.0500 \, L \), \( M_{H_2SO_4} = 0.50 \, M \).
\( n_{H_2SO_4} = 0.50 \, mol/L \times 0.0500 \, L = 0.025 \, mol \).

Step2: Calculate moles of \( NaOH \)

\( V_{NaOH} = 60.0 \, mL = 0.0600 \, L \), \( M_{NaOH} = 0.80 \, M \).
\( n_{NaOH} = 0.80 \, mol/L \times 0.0600 \, L = 0.048 \, mol \).

Step3: Determine limiting reactant using stoichiometry

From reaction \( 2NaOH + H_2SO_4
ightarrow 2H_2O + Na_2SO_4 \), mole ratio \( NaOH:H_2SO_4 = 2:1 \).
Moles of \( NaOH \) needed for \( 0.025 \, mol \, H_2SO_4 \): \( 2 \times 0.025 = 0.050 \, mol \).
But we have \( 0.048 \, mol \, NaOH \) (less than 0.050), so \( NaOH \) is limiting.

Step4: Calculate moles of \( Na_2SO_4 \) from limiting reactant

Mole ratio \( NaOH:Na_2SO_4 = 2:1 \).
\( n_{Na_2SO_4} = \frac{n_{NaOH}}{2} = \frac{0.048}{2} = 0.024 \, mol \).
(Check with \( H_2SO_4 \): ratio \( H_2SO_4:Na_2SO_4 = 1:1 \), \( n_{Na_2SO_4} = 0.025 \, mol \), but limited by \( NaOH \), so use \( NaOH \)'s moles.)

Answer:

\( 0.024 \)