Question

One process for the commercial production of baking soda (sodium hydrogen carbonate) involves the following reaction, in which the carbon dioxide is used in its solid form ("dry ice" enough for the sodium hydrogen carbonate to precipitate: $$ \mathrm{NaCl}(a q)+\mathrm{NH}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(s) \rightarrow \mathrm{NH}_{4} \mathrm{Cl}(a q)+\mathrm{NaHCO}_{3}(s) $$ Because they are relatively cheap, sodium chloride and water are typically present in excess. What is the expected yield of \(\mathrm{NaHCO}_{3}\) when one performs such a synthesis using \(10.0 \mathrm{~g}\) of ammonia and \(15.0 \mathrm{~g}\) of dry ice, with an excess of \(\mathrm{NaCl}\) and water?

Short answer

The expected yield of sodium hydrogen carbonate (NaHCO3) when 10.0 grams of ammonia and 15.0 grams of dry ice are used in this synthesis process, with an excess of NaCl and water, is approximately 28.6 grams.

Step-by-step solution

Find the molar masses

To convert the mass of the reactants to moles, we first need to determine their molar masses. Look up the atomic masses of each element and use them to find the molar masses of ammonia (NH3) and carbon dioxide (CO2). The molar mass of NH3 is \( 14.01\ (\text{N}) + 3 \times 1.01\ (\text{H}) = 17.03\ \text{g/mol}\), and the molar mass of CO2 is \(12.01\ (\text{C}) + 2 \times 16.00\ (\text{O}) = 44.01\ \text{g/mol}\).

Calculate moles of reactants

Now we can convert the given masses of ammonia and dry ice to moles using the molar masses we found. For ammonia, moles = \(\frac{10.0\ \text{g}}{17.03\ \text{g/mol}} = 0.587\ \text{mol}\) For dry ice, moles = \(\frac{15.0\ \text{g}}{44.01\ \text{g/mol}} = 0.341\ \text{mol}\) #Step 2: Identify the limiting reactant#

Compare molar ratios

According to the balanced chemical equation, 1 mole of ammonia (NH3) and 1 mole of dry ice (CO2) react together to produce 1 mole of NaHCO3. Now, we compare the molar ratios between ammonia and dry ice to the stoichiometric ratio in the balanced equation. Ammonia: \(\frac{0.587\ \text{mol}}{1} = 0.587\) Dry ice: \(\frac{0.341\ \text{mol}}{1} = 0.341\) Since the ratio for dry ice (0.341) is smaller than the ratio for ammonia (0.587), dry ice (CO2) is the limiting reactant. #Step 3: Calculate the expected yield of NaHCO3#

Use stoichiometry

As the limiting reactant is carbon dioxide, we use its moles to calculate the moles of sodium hydrogen carbonate (NaHCO3) produced according to the stoichiometry of the balanced chemical equation (1:1 ratio): Moles of NaHCO3 produced = \(0.341\ \text{mol}\) Now, we can convert the moles of NaHCO3 back into grams using its molar mass. Molar mass of NaHCO3 = \(22.99\ (\text{Na}) + 1.01\ (\text{H}) + 12.01\ (\text{C}) + 3 \times 16.00\ (\text{O}) = 84.01\ \text{g/mol}\) Expected yield of NaHCO3 = \(0.341\ \text{mol} \times 84.01\ \text{g/mol} = 28.6\ \text{g}\) The expected yield of sodium hydrogen carbonate (NaHCO3) when 10.0 grams of ammonia and 15.0 grams of dry ice are used in this synthesis process, with an excess of NaCl and water, is approximately 28.6 grams.

Key concepts

Limiting Reactant

Understanding the role of the limiting reactant in a chemical reaction is crucial as it determines how much product can be formed. It's the substance that's consumed first, thus limiting the amount of product produced. To identify it, we compare the mole ratio of the reactants used to the mole ratio from the balanced equation.

When given amounts of two or more reactants, first convert these quantities to moles. Then, find the ratio of the moles actually used to the moles required by the equation. The reactant with the smallest ratio – that runs out first – is your limiting reactant. In the provided example, carbon dioxide (CO2) is the limiting reactant because it has a smaller ratio compared to ammonia (NH3). This tells us that CO2 will determine the maximum amount of sodium hydrogen carbonate (NaHCO3) that can be produced.

Molar Mass

Molar mass is a fundamental concept in stoichiometry, representing the mass of one mole of a substance, measured in grams per mole (g/mol). It's calculated by adding up the atomic masses of each element in a compound, found on the periodic table. As seen in our exercise, knowing the molar masses of NH3 and CO2 allows us to convert the mass of each reactant into moles. This step is critical because reactions happen on a molecular level, and stoichiometry requires balancing of moles, not mass.

Once you've determined the molar mass, use it as a conversion factor to translate grams into moles or vice versa. This conversion helps in both finding the limiting reactant and calculating the theoretical yield of a product, making it a vital step in solving stoichiometry problems.

Theoretical Yield

The theoretical yield is the maximum amount of product that can be generated as predicted by the chemical equation, assuming everything reacts perfectly. It's a key concept for efficiency and planning in chemical synthesis.

To calculate the theoretical yield, you must identify the limiting reactant and then use stoichiometry to determine how many moles of product it could produce. This quantity is then converted into grams using the product's molar mass. As shown in the example, using the molar mass of NaHCO3 and the moles of the limiting reactant, we can predict the maximum possible yield. Here, the calculation indicated an expected yield of 28.6 grams. Remember, in reality, the actual yield is often less than the theoretical yield due to inefficiencies, side reactions, or incomplete reactions.

Chemical Reaction

A chemical reaction involves the transformation of one or more substances into new substances. It is represented by a balanced chemical equation that shows the reactants on the left and the products on the right. Stoichiometry comes into play as it allows us to calculate just how much of each substance is needed or produced during the reaction.

Having a balanced equation ensures that mass is conserved, based on the Law of Conservation of Mass. This means all atoms present in the reactants must be accounted for in the products. In the exercise, the balanced equation helps us understand the stoichiometry – each mole of ammonia and carbon dioxide produces one mole of NaHCO3 and NH4Cl, illustrating a 1:1 ratio necessary for these computations.