Question

Sodium hydroxide reacts with carbon dioxide as follows: $$ 2 \mathrm{NaOH}(s)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) $$ Which is the limiting reactant when \(1.85 \mathrm{~mol} \mathrm{NaOH}\) and \(1.00 \mathrm{~mol} \mathrm{CO}_{2}\) are allowed to react? How many moles of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) can be produced? How many moles of the excess reactant remain after the completion of the reaction?

Short answer

The limiting reactant is Sodium hydroxide (NaOH). 0.925 moles of Sodium carbonate (Na2CO3) can be produced. 0.075 moles of Carbon dioxide (CO2) remain after the reaction is complete.

Step-by-step solution

Determine the mole ratio

To find the mole ratio, divide the given moles of each reactant by their respective coefficients in the balanced equation: Mole ratio of NaOH = \(\frac{1.85}{2}\) Mole ratio of CO2 = \(\frac{1.00}{1}\)

Identify the limiting reactant

The reactant with the lowest mole ratio in step 1 will be the limiting reactant. In this case: Mole ratio of NaOH = 0.925 Mole ratio of CO2 = 1.00 Since 0.925 < 1.00, NaOH is the limiting reactant.

Calculate the moles of Na2CO3 produced

Using the stoichiometric ratio from the balanced equation, we can calculate the moles of Na2CO3 produced by the limiting reactant (NaOH): \(moles\, of\, Na_{2}CO_{3} = moles\, of\, NaOH \cdot \frac{1 \,mol\, Na_{2}CO_{3}}{2 \,mol\, NaOH} = 1.85 \,mol\, NaOH \cdot \frac{1}{2} = 0.925 \,mol\, Na_{2}CO_{3}\)

Calculate the remaining moles of the excess reactant

To find the remaining moles of the excess reactant (CO2) after the reaction, we first need to calculate how many moles of CO2 reacted with the limiting reactant (NaOH): \(moles\, of\, CO_{2}\, reacted = moles\, of\, NaOH \cdot \frac{1 \,mol\, CO_{2}}{2 \,mol\, NaOH} = 1.85 \,mol\, NaOH \cdot \frac{1}{2} = 0.925 \,mol\, CO_{2}\) Now, subtract the moles of CO2 reacted from the initial moles of CO2: \(remaining\, moles\, of\, CO_{2} = initial\, moles\, of\, CO_{2} - moles\, of\, CO_{2}\, reacted = 1.00 \,mol\, CO_{2} - 0.925 \,mol\, CO_{2} = 0.075 \,mol\, CO_{2}\) To summarize: 1. The limiting reactant is Sodium hydroxide (NaOH). 2. 0.925 moles of Sodium carbonate (Na2CO3) can be produced. 3. 0.075 moles of Carbon dioxide (CO2) remain after the reaction is complete.

Key concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that deals with the calculation of reactants and products in chemical reactions. At its core, stoichiometry allows chemists to predict the amount of substances consumed and produced in a given reaction. This is based on the balanced chemical equation, which provides the mole ratio of reactants and products. In our exercise, the balanced equation is:\[2 \mathrm{NaOH}(s) + \mathrm{CO}_{2}(g) \rightarrow \mathrm{Na}_{2}\mathrm{CO}_{3}(s) + \mathrm{H}_{2} \mathrm{O}(l)\]From this equation, we can see that 2 moles of sodium hydroxide (NaOH) react with 1 mole of carbon dioxide (CO2) to produce 1 mole of sodium carbonate (Na2CO3) and 1 mole of water (H2O). By understanding this mole ratio, you can determine which reactant will be used up first and thus is the limiting reactant. Knowing the limiting reactant is crucial because it determines the maximum amount of product that can be formed in the reaction.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products, through the breaking and forming of chemical bonds. In our given reaction, sodium hydroxide (NaOH) reacts with carbon dioxide (CO2) to form sodium carbonate (Na2CO3) and water (H2O). This type of reaction can be classified as a synthesis reaction, where two reactants combine to form a single more complex product. Understanding chemical reactions requires recognizing how reactants transform into products through a series of steps:

• Identify the reactants and write out the balanced chemical equation.
• Use the equation to understand the stoichiometric relationships between reactants and products.
• Determine the limiting reactant to predict how much product will be formed and which reactants are in excess.

This particular chemical reaction involves a solid (NaOH) reacting with a gas (CO2) and results in the formation of a solid (Na2CO3) and a liquid (H2O). It illustrates the importance of balanced equations in predicting the outcomes and quantities involved in chemical reactions.

Mole Calculations

Mole calculations are critical in chemistry as they provide a means to quantify the amount of substance involved in a chemical reaction. A mole is a standard unit in chemistry that is used to express amounts of a chemical substance. It allows for a bridge between the atomic scale and real-world scale.In our exercise, mole calculations help us to determine the amount of sodium carbonate (Na2CO3) produced and the remaining quantities of the excess reactant. Let's break down the calculations for clarity:

• Firstly, calculate the mole ratio of each reactant from the balanced equation. For NaOH, the mole ratio is \(\frac{1.85}{2} = 0.925\). For CO2, it is \(1.00\).
• Identify the limiting reactant, which is NaOH, as it has the smaller mole ratio.
• Calculate the moles of Na2CO3 produced using the mole ratio from the limiting reactant: \(moles\, of\, Na_{2}CO_{3} = 1.85 \times \frac{1}{2} = 0.925\, \text{mol}\).
• Determine the remaining moles of the excess reactant (CO2): \(1.00 \text{ mol CO}_2 - 0.925 \text{ mol CO}_2 = 0.075\text{ mol CO}_2\).

Through mastering these mole calculations, you'll be better equipped to predict the amount of product formed and understand the quantities of reactants involved in any chemical reaction.