I have to find the theoretical amount of sodium bicarbonate. The steps I followed in order to produce it are:

  1. 15g $\ce{NaCl}$ are added to a beaker. Then we add 87.5 mL of $\ce{NH3}(12M)$

  2. We add 115g of dry ice(whose density is 1.5 g/mL)

  3. Then the solid is taken with filtration and we rinse it first with 9mL of cold water and then with 6mL of acetone.

Then we put the solid in the oven. I know that the product after we take it off the oven will be sodium bicarbonate. I believe that in order to find the theoretical amount of sodium bicarbonate I have to write down the reactions that are taking place.

$\ce{NaCl + NH_3 + CO_2 + H_2O → NH_4Cl + NaHCO_3}$

But I'm confused as to if there is another reaction where $\ce{NH_4Cl}$ and $\ce{NaHCO_3}$ react with water or acetone. Also, working with the stoichiometry of the reactions, from which reactant is it better to conclude about the moles of NaHCO3?

  • $\begingroup$ considering your second question, when looking at the reaction $\ce{2A + B -> AB + A}$, how much AB do you think you get? 1 or 2? $\endgroup$ – Fl.pf. Mar 27 '17 at 10:12
  • $\begingroup$ I get 1 AB. But we have to know the moles of A and B in order to see how much reacts from each and how much AB we will finally get. $\endgroup$ – Maria P Mar 27 '17 at 10:16
  • $\begingroup$ yeah. So have you calculated how many moles you have of your reactants? $\endgroup$ – Fl.pf. Mar 27 '17 at 10:22
  • $\begingroup$ The moles of NaCl are 0.256, the moles of NH3 are 1.05. I'm confused about the moles of CO2(which is produced by the dry ice) $\endgroup$ – Maria P Mar 27 '17 at 10:31
  • $\begingroup$ $\ce{CO_2}$ moles are not produced by dry ice. Dry ice is $\ce{CO_2}$. You need 1 $\ce{NaCl}$ to get 1 $\ce{NaHCO_3}$, right? So if you have 0.256 mol of $\ce{Na}$, how much $\ce{NaHCO_3}$ could you theoretically synthesize (no losses, no by-products)? $\endgroup$ – Fl.pf. Mar 27 '17 at 10:34

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