The solvay process in an industrial method for preparing baking soda. In the process, CO2, NH3, H2O, and NaCl react to produce NaHCO3. If 25.0 L of carbon dioxide gas at STP and 20.0 L of ammonia gas at STP react in excess water and salt, what is the mass of baking soda produced ? The equation for the reaction is: CO2 (g) + NH3(g)+H2O(l)+NaCl(s)= NaHCO3(s)+NH4Cl(aq)
CO2 x (1 mol/22.4L) = 1.12 mols CO2.
NH3 x (1 mol/22.4L) = 0.89 mols NH3.
Therefore, NH3 is the limiting reagent and you will produce 0.89 mol NaHCO3. Convert that to grams.
0.89 mol NaHCO3* 84.01g = 74.8 g ?
For the following reaction, 0.431 moles of iron are mixed with 0.293 moles of oxygen gas.
iron (s) + oxygen (g) iron(II) oxide (s)
What is the FORMULA for the limiting reagent
75.0 g NaHCO3
To find the mass of baking soda produced in the Solvay process, we need to use stoichiometry. Stoichiometry is a relationship between the quantities of reactants and products in a chemical reaction.
First, we need to determine the balanced equation for the reaction. From the given information, we have:
CO2 (g) + NH3(g) + H2O(l) + NaCl(s) = NaHCO3(s) + NH4Cl(aq)
Next, we need to convert the volume of each gas at STP (Standard Temperature and Pressure) to moles using the ideal gas law. At STP, 1 mole of any gas occupies 22.4 liters of volume.
For carbon dioxide (CO2):
1 mole CO2 = 22.4 L CO2
25.0 L CO2 = 25.0 L CO2 x (1 mole CO2 / 22.4 L CO2)
For ammonia (NH3):
1 mole NH3 = 22.4 L NH3
20.0 L NH3 = 20.0 L NH3 x (1 mole NH3 / 22.4 L NH3)
Now that we have the moles of carbon dioxide and ammonia, we can determine which reactant is limiting and will determine the amount of product formed. The balanced equation gives us the stoichiometric ratio between the reactants and the product:
CO2 : NH3 : NaHCO3
1 : 1 : 1
Comparing the moles of CO2 and NH3, we can see that they have an equal ratio. This means neither is the limiting reactant; both reactants will be completely consumed.
The moles of CO2 and NH3 equal the moles of NaHCO3 produced.
Next, we need to convert the moles of NaHCO3 to grams using its molar mass. The molar mass of NaHCO3 is:
Na = 22.99 g/mol
H = 1.01 g/mol
C = 12.01 g/mol
O = 16.00 g/mol
NaHCO3 = 22.99 + 1.01 + 12.01 + (3 x 16.00) g/mol
Now, we can calculate the mass of baking soda (NaHCO3) produced:
Mass of NaHCO3 = moles of NaHCO3 x molar mass of NaHCO3
Note: Since the reaction takes place in excess water and salt (NaCl), we can assume all of the reactants will be consumed.
If you provide the final moles of CO2 and NH3, I can help you calculate the mass of baking soda produced.