Limiting Reagent

Question

Mixture of 0.020 mol NaOH and 0.015 mol H2SO4 reacts (complete neutralization). Moles of salt formed (Na2SO4) are:

RankGuru · Accepted Answer

1. **Balanced Equation**: $2NaOH + H_2SO_4 	o Na_2SO_4 + 2H_2O$.
2. **Stoichiometry Ratio**: $2$ moles of $NaOH$ react with $1$ mole of $H_2SO_4$ to produce $1$ mole of $Na_2SO_4$.
3. **Identify Limiting Reagent**:
   - To react with $0.015$ mol $H_2SO_4$, we would need $2 	imes 0.015 = 0.030$ mol $NaOH$.
   - We only have $0.020$ mol $NaOH$.
   - Therefore, **$NaOH$ is the limiting reagent**.
4. **Calculate Product**: 
   - Based on $NaOH$: $2$ moles $NaOH 	o 1$ mole $Na_2SO_4$.
   - $0.020$ moles $NaOH 	o \frac{1}{2} 	imes 0.020 = 0.010$ mol $Na_2SO_4$.
5. **Result**: **0.010 moles** of salt are formed.

Limiting Reagent

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