NaCl (aq) + NH3(aq) + H2O(l) + CO2(s) --> NH4Cl(aq) + NaHCO3(s)

because they are relatively cheap, sodium chloride and water are typically present in excess. what is the expected yield of NaHCO3 when one performs such a synthesis using 10.0 g of ammonia and 15.0 g of dry ice - CO2(s), with an excess of NaCl and water-H20?

Convert 10.0 g NH3 to moles. moles = grams/molar mass.

Convert 15.0 g CO2 to moles.

Now convert moles NH3 to moles product and convert moles CO2 to moles product. Determine the limiting reagent. Finally, convert moles of the product to grams product.

To find the expected yield of NaHCO3, we need to determine the limiting reactant in the reaction. The limiting reactant is the one that will be completely consumed and, therefore, determines the maximum amount of product that can be formed.

First, we need to calculate the number of moles of each reactant using their molar masses.

Molar mass of ammonia (NH3) = 17.03 g/mol
Molar mass of dry ice (CO2) = 44.01 g/mol

moles of NH3 = mass / molar mass = 10.0 g / 17.03 g/mol = 0.587 mol
moles of CO2 = mass / molar mass = 15.0 g / 44.01 g/mol = 0.341 mol

Next, we can write a balanced chemical equation to determine the stoichiometry of the reaction:

2NaCl(aq) + 2NH3(aq) + H2O(l) + CO2(s) → 2NH4Cl(aq) + NaHCO3(s)

From the balanced equation, we see that the stoichiometric ratio between NH3 and NaHCO3 is 2:1. This means that for every 2 moles of NH3, we can expect 1 mole of NaHCO3.

Since the stoichiometric ratio is 2:1, the number of moles of NH3 will be twice as much as the number of moles of NaHCO3 formed.

moles of NaHCO3 = 0.587 mol NH3 / 2 = 0.2935 mol

Now, we can calculate the mass of NaHCO3 using its molar mass:

Molar mass of NaHCO3 = 84.01 g/mol

mass of NaHCO3 = moles x molar mass = 0.2935 mol x 84.01 g/mol = 24.65 g

Therefore, the expected yield of NaHCO3 is 24.65 grams when 10.0 grams of ammonia and 15.0 grams of dry ice are used, assuming an excess of NaCl and water.