Question

h₂so₄ + 2koh → k₂so₄ + 2h₂o
30.0 ml of 1.30 m h₂so₄ is added to
25.0 ml of 2.00 m koh. the resulting
temperature change is an increase of
9.55 °c.

cₛₒₗ = 4.20 j/g °c dₛₒₗ = 1.00 g/ml
massₛₒₗ = 55.0 g qᵣₓₙ = −2,210 j
how many moles of k₂so₄ will form?
molesᵣₓₙ = ? mol

Explanation:

Step1: Calculate moles of \(H_2SO_4\)

Molarity formula: \(n = M \times V\) (in liters).
\(V_{H_2SO_4} = 30.0\space mL = 0.0300\space L\), \(M_{H_2SO_4} = 1.30\space M\).
\(n_{H_2SO_4} = 1.30\space mol/L \times 0.0300\space L = 0.0390\space mol\).

Step2: Calculate moles of \(KOH\)

\(V_{KOH} = 25.0\space mL = 0.0250\space L\), \(M_{KOH} = 2.00\space M\).
\(n_{KOH} = 2.00\space mol/L \times 0.0250\space L = 0.0500\space mol\).

Step3: Determine limiting reactant

Reaction: \(H_2SO_4 + 2KOH
ightarrow K_2SO_4 + 2H_2O\).
Mole ratio \(H_2SO_4:KOH = 1:2\).
For \(0.0390\space mol\space H_2SO_4\), required \(KOH = 0.0390 \times 2 = 0.0780\space mol\) (exceeds available \(0.0500\space mol\)).
For \(0.0500\space mol\space KOH\), required \(H_2SO_4 = \frac{0.0500}{2} = 0.0250\space mol\) (less than available \(0.0390\space mol\)).
Thus, \(KOH\) is limiting.

Step4: Calculate moles of \(K_2SO_4\)

From reaction, \(2\space mol\space KOH\) produces \(1\space mol\space K_2SO_4\).
Moles of \(K_2SO_4 = \frac{0.0500\space mol\space KOH}{2} = 0.0250\space mol\).

Answer:

\(0.0250\)