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In chemical reactions, the limiting reactant is the substance that is completely consumed first, thus determining the amount of products formed. It essentially limits the extent of the reaction because once it is used up, the reaction cannot proceed further. Identifying the limiting reactant is crucial in stoichiometry, as it helps in calculating the maximum yield of products.
To find the limiting reactant in a reaction like the one involving sulfur dioxide (SOâ‚‚) and sodium hydroxide (NaOH), we compare the mole ratio of the reactants as given by the balanced chemical equation. The balanced equation for the given reaction is:
\[\mathrm{SO}_{2}(g) + 2 \mathrm{NaOH}(s) \rightarrow \mathrm{Na}_{2}\mathrm{SO}_{3}(s) + \mathrm{H}_{2}\mathrm{O}(l)\]

• According to the equation, 1 mole of SOâ‚‚ reacts with 2 moles of NaOH.
• In our calculations, we had 0.597 mol of SOâ‚‚ and 0.820 mol of NaOH.
• The actual mole ratio is 0.597:0.820, or 1:1.373, which is less than the required 1:2.

This means SOâ‚‚ is the limiting reactant because it is present in the lesser proportion than required. Identifying this helps us determine the exact amount of product that can be formed.

Balancing a chemical equation is a critical step in stoichiometry to ensure that the law of conservation of mass is satisfied. This states that matter cannot be created or destroyed in a closed system. Therefore, the number of each type of atom on the reactant side must equal the number on the product side.
Let's consider our chemical equation:

• Unbalanced equation: \( \mathrm{SO}_{2}(g) + \mathrm{NaOH}(s) \rightarrow \mathrm{Na}_{2}\mathrm{SO}_{3}(s) + \mathrm{H}_{2} \mathrm{O}(l) \).
• Balanced equation: \( \mathrm{SO}_{2}(g) + 2 \mathrm{NaOH}(s) \rightarrow \mathrm{Na}_{2}\mathrm{SO}_{3}(s) + \mathrm{H}_{2}\mathrm{O}(l) \).

By adding a coefficient of 2 in front of NaOH, we balanced two sodium (Na), four oxygen (O), two hydrogen (H), and one sulfur (S) atom on both sides of the equation. This balanced equation helps us use stoichiometric coefficients to convert between moles of reactants and products.

Converting between moles and mass is an essential skill in stoichiometry, allowing us to quantify the substances involved in a chemical reaction. To perform a mole-to-mass conversion, we use the molar mass of the substance, which is the mass of one mole of that substance.
Here's how we apply it:

• Find the molar mass of each reactant or product. For example, SOâ‚‚ has a molar mass of 64.1 g/mol, and NaOH has a molar mass of 39.997 g/mol.
• Use the molar mass to convert grams to moles or vice versa. The formula is: \[\text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}\]
• For instance, 38.3 g of SOâ‚‚ is equivalent to \(\frac{38.3}{64.1} \approx 0.597 \text{ moles} \).

When converting products' moles back into mass, the same molar mass principle applies. This step is crucial for determining how much of each product is formed from a given amount of reactants. By completing these conversions, we bridge the macroscopic amounts we can measure with the microscopic amounts involved in reactions.