Question

2naoh + h₂so₄ → na₂so₄ + 2h₂o
13.2 g naoh and 13.2 g h₂so₄ react.
the limiting reactant is:
○ na₂so₄
○ naoh
○ h₂o
○ h₂so₄

Explanation:

Step1: Calculate moles of NaOH

Molar mass of NaOH: \(M_{NaOH}=40\space g/mol\)
Moles of NaOH: \(n_{NaOH}=\frac{m}{M}=\frac{13.2\space g}{40\space g/mol}=0.33\space mol\)

Step2: Calculate moles of \(H_2SO_4\)

Molar mass of \(H_2SO_4\): \(M_{H_2SO_4}=98\space g/mol\)
Moles of \(H_2SO_4\): \(n_{H_2SO_4}=\frac{m}{M}=\frac{13.2\space g}{98\space g/mol}\approx0.135\space mol\)

Step3: Use stoichiometry

From the balanced equation \(2NaOH + H_2SO_4
ightarrow Na_2SO_4+2H_2O\), the mole ratio of \(NaOH:H_2SO_4 = 2:1\)
If all \(H_2SO_4\) reacts, moles of NaOH needed: \(n_{NaOH\space needed}=2\times n_{H_2SO_4}=2\times0.135 = 0.27\space mol\)
Since \(0.33\space mol\) (available NaOH) \(>0.27\space mol\) (needed NaOH) when \(H_2SO_4\) is fully consumed.
If all NaOH reacts, moles of \(H_2SO_4\) needed: \(n_{H_2SO_4\space needed}=\frac{n_{NaOH}}{2}=\frac{0.33}{2}=0.165\space mol\)
Since \(0.135\space mol\) (available \(H_2SO_4\)) \(<0.165\space mol\) (needed \(H_2SO_4\)) when NaOH is fully consumed.

Answer:

\(H_2SO_4\)