If $1 ext{ gram}$ of sodium hydroxide was treated with $25 ext{ mL}$ of $0.75 ext{ M}$ HCl solution, the

Question

If $1 	ext{ gram}$ of sodium hydroxide was treated with $25 	ext{ mL}$ of $0.75 	ext{ M}$ HCl solution, the mass of sodium hydroxide left unreacted is equal to:

Accepted Answer

1. **Calculate Moles of Reactants:**
   - Moles of NaOH ($n_{	ext{NaOH}}$) = $\frac{	ext{Mass}}{	ext{Molar Mass}} = \frac{1 	ext{ g}}{40 	ext{ g/mol}} = 0.025 	ext{ mol}$.
   - Moles of HCl ($n_{	ext{HCl}}$) = $	ext{Molarity} 	imes 	ext{Volume (L)}$ [Class 12 Chemistry, Eq. 1.8].
     $$ n_{	ext{HCl}} = 0.75 	ext{ mol L}^{-1} 	imes 0.025 	ext{ L} = 0.01875 	ext{ mol} $$
2. **Determine Limiting Reagent:**
   - The balanced reaction is: $	ext{NaOH} + 	ext{HCl} ightarrow 	ext{NaCl} + 	ext{H}_2	ext{O}$.
   - Stoichiometry is 1:1. Since $0.01875 	ext{ mol} < 0.025 	ext{ mol}$, HCl is the limiting reagent.
3. **Calculate Unreacted NaOH:**
   - Moles consumed = Moles of HCl = $0.01875 	ext{ mol}$.
   - Moles remaining = $0.025 - 0.01875 = 0.00625 	ext{ mol}$.
4. **Convert to Mass:**
   - Mass = $	ext{Moles} 	imes 	ext{Molar Mass} = 0.00625 	ext{ mol} 	imes 40 	ext{ g/mol} = 0.25 	ext{ g}$.
   - Convert to mg: $0.25 	ext{ g} 	imes 1000 	ext{ mg/g} = 250 	ext{ mg}$.

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