Question

The Solvay process is used to manufacture baking soda, NaHCO3. In the process, CO2, NH3, HzO, and NaCl react to produce baking soda. If 15.0 L CO2 and 10.0 L NH3 react at STP, with excess water and sodium chloride, what is the limiting reactant

Answer 1

The limiting reactant between is NH₃.

Step-by-step explanation:

The reaction of the Solvay process is:

CO₂(g) + NH₃(g) + H₂O(l) + NaCl(s) ⇄ NaHCO₃(s) + NH₄Cl(aq) (1)

Since the water and the sodium chloride are in excess we need to find the number of moles of CO₂ and NH₃ at STP (1 amt, 273 K).

`PV = nRT \rightarrow n = (PV)/(RT)`

Where:

n: is the number of moles

P: is the pressure = 1 atm

V: is the volume

T: is the temperature = 273 K

R: is the gas constant = 0.082 L*atm(K*mol)

For CO₂ we have:

`n = (PV)/(RT) = (1 atm*15.0 L)/(0.082 L*atm/(K*mol)*273 K) = 0.67 moles`

And for NH₃ we have:

`n = (PV)/(RT) = (1 atm*10.0 L)/(0.082 L*atm/(K*mol)*273 K) = 0.45 moles`

From the equation (1) we have that 1 mol of CO₂ reacts with 1 mol of NH₃, so from that ratio we have:

`n_{CO_(2)} = \frac{\eta_{CO_(2)}}{\eta_{NH_(3)}}*n_{NH_(3)} = (1)/(1)*0.45 moles} = 0.45 moles`

From above we have that 1 mol of NH₃ reacts with 0.45 moles of CO₂, and we have 0.67 moles of CO₂, hence the limiting reactant is NH₃.

Therefore, the limiting reactant between CO₂ and NH₃ is NH₃.

I hope it helps you!