If you have 5.8 g of NaOH and 4.2g of HCl which will be your limiting reagent in the equation NaOH + HCl to NaCl + H2O

To find the limiting reagent, we need to compare the moles of each reactant based on the balanced chemical equation:

\[
\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}
\]

This equation tells us that one mole of NaOH reacts with one mole of HCl. Now, we need to calculate the number of moles for both NaOH and HCl.

### Step 1: Molar Masses
- **Molar mass of NaOH (sodium hydroxide)**:
\[
\text{Na} (23.0 \, \text{g/mol}) + \text{O} (16.0 \, \text{g/mol}) + \text{H} (1.0 \, \text{g/mol}) = 40.0 \, \text{g/mol}
\]

- **Molar mass of HCl (hydrochloric acid)**:
\[
\text{H} (1.0 \, \text{g/mol}) + \text{Cl} (35.5 \, \text{g/mol}) = 36.5 \, \text{g/mol}
\]

### Step 2: Moles of Each Reactant
- **Moles of NaOH**:
\[
\frac{5.8 \, \text{g}}{40.0 \, \text{g/mol}} = 0.145 \, \text{mol}
\]

- **Moles of HCl**:
\[
\frac{4.2 \, \text{g}}{36.5 \, \text{g/mol}} = 0.115 \, \text{mol}
\]

### Step 3: Identify the Limiting Reagent
Since both reactants react in a 1:1 molar ratio, the reactant with fewer moles will be the limiting reagent. In this case, HCl has fewer moles (0.115 mol) than NaOH (0.145 mol).

**Conclusion**: **HCl** is the limiting reagent.

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Would you like further details or have any questions?

### Related questions:
1. How do you determine the excess reagent after identifying the limiting reagent?
2. What happens if you add more NaOH to the reaction?
3. How does the concept of molarity relate to limiting reagents in a reaction?
4. How would this reaction change if the quantities of NaOH and HCl were reversed?
5. What is the total mass of the products in this reaction?

**Tip**: Always balance the chemical equation first before identifying limiting reagents in any reaction.

Find out how to identify the limiting reagent in the reaction between 5.8 g of NaOH and 4.2 g of HCl. This problem involves key concepts like stoichiometry, molar mass calculations, and limiting reagents. Detailed explanation with step-by-step solution, ideal for Grades 10-12.

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