Question

h₂so₄ + 2koh → k₂so₄ + 2h₂o
50.0 ml of 2.00 m h₂so₄ is added to
50.0 ml of 2.00 m koh. the resulting
temperature change is an increase of
14.5 °c.

c_sol = 4.20 j/g °c d_sol = 1.00 g/ml c_cal = 5.70 j/°c
mass_soln = 100. g q_rxn = -6,170 j

how many moles of k₂so₄ will form?

moles_rxn = ? mol

Explanation:

Step1: Calculate moles of H₂SO₄

Molarity \( M = \frac{\text{moles}}{\text{volume (L)}} \), so moles of \( \text{H}_2\text{SO}_4 = M \times V \).
Volume of \( \text{H}_2\text{SO}_4 = 50.0 \, \text{mL} = 0.0500 \, \text{L} \), \( M = 2.00 \, \text{M} \).
\( \text{moles of } \text{H}_2\text{SO}_4 = 2.00 \, \text{mol/L} \times 0.0500 \, \text{L} = 0.100 \, \text{mol} \).

Step2: Calculate moles of KOH

Volume of \( \text{KOH} = 50.0 \, \text{mL} = 0.0500 \, \text{L} \), \( M = 2.00 \, \text{M} \).
\( \text{moles of } \text{KOH} = 2.00 \, \text{mol/L} \times 0.0500 \, \text{L} = 0.100 \, \text{mol} \).

Step3: Determine limiting reactant

From the reaction \( \text{H}_2\text{SO}_4 + 2\text{KOH}
ightarrow \text{K}_2\text{SO}_4 + 2\text{H}_2\text{O} \), the mole ratio of \( \text{H}_2\text{SO}_4 : \text{KOH} \) is \( 1:2 \).
For 0.100 mol \( \text{H}_2\text{SO}_4 \), required \( \text{KOH} = 0.100 \times 2 = 0.200 \, \text{mol} \), but we have only 0.100 mol \( \text{KOH} \). Thus, \( \text{KOH} \) is limiting.

Step4: Calculate moles of \( \text{K}_2\text{SO}_4 \) from limiting reactant

Mole ratio of \( \text{KOH} : \text{K}_2\text{SO}_4 = 2:1 \).
\( \text{moles of } \text{K}_2\text{SO}_4 = \frac{\text{moles of } \text{KOH}}{2} = \frac{0.100 \, \text{mol}}{2} = 0.0500 \, \text{mol} \).

Answer:

0.0500