2) Given the balanced reaction equation predict the moles of each product given the following reactants:
AlCl3 + 3 NaHCO3 Al(OH)3 + 3 NaCl + 3 CO2
AlCl3 NaHCO3 Al(OH)3 NaCl 58.4 CO2
a 1 mol 1 mol ? ?
b 1 mol 3 mol ? ?
c 3 mol 1 mol ? ?
These are limiting reagent problems. I solve these the long way. Starting with one of the reactants, calculate how much of one of the products will be formed. Do the same for the other reactant and the same product. If both answers are the same everything is in the right proportion and there is no limiting reagent(LR). If not, the SMALLER value of the product will be formed (and the reagent producing that smaller value will be the limiting reagent). First one:
...AlCl3+3NaHCO3 =>Al(OH)3+3NaCl+3CO2
...1 mol...1 mol.....x......x......x
1 mol AlCl3 will produce3 1 mol Al(OH)3.
1 mol NaHCO3 will produce 1/3 mol Al(OH)3 so NaHCO3 is the limiting reagent. You will produce 1/3 mol Al(OH)3. At the same time you will produce 1 mol NaCl and 1 mol CO2
b part:
....1......3.........x.......x.....x
1 mol AlCl3 will produce 1 mol Al(OH)3
3 mol NaHCO3 will produce 1 mol Al(OH)3 which means those are the exact proportions you need and there is no LR. You will produce 3 mol NaCl and 3 mols CO2 at the same time you produce 1 mol Al(OH)3.
I will leave c part for you. Post your work if you want me to check it.
Thank you very much for your time on this and the other problems I posted. You are appreciated.
Selina Olomua
To predict the moles of each product given the reactants, you need to use the coefficients in the balanced equation as conversion factors.
The balanced equation is: AlCl3 + 3 NaHCO3 -> Al(OH)3 + 3 NaCl + 3 CO2
From the equation, you can see that 1 mole of AlCl3 reacts with 3 moles of NaHCO3 to produce 1 mole of Al(OH)3, 3 moles of NaCl, and 3 moles of CO2.
Now let's look at the options provided:
a) 1 mole of AlCl3 and 1 mole of NaHCO3:
According to the balanced equation, for every 1 mole of AlCl3, 1 mole of NaHCO3 is required. Therefore, you would predict that the reaction would produce 1 mole of Al(OH)3, 3 moles of NaCl, and 3 moles of CO2.
b) 1 mole of AlCl3 and 3 moles of NaHCO3:
Again, according to the balanced equation, for every 1 mole of AlCl3, 3 moles of NaHCO3 are required. In this case, you would predict that the reaction would produce 1 mole of Al(OH)3, 3 moles of NaCl, and 3 moles of CO2, just like in option a.
c) 3 moles of AlCl3 and 1 mole of NaHCO3:
Here, for every 3 moles of AlCl3, 1 mole of NaHCO3 is required according to the balanced equation. Therefore, you would predict that the reaction would produce an unknown amount of Al(OH)3, 3 moles of NaCl, and 3 moles of CO2.
So in conclusion, option a and b would produce the same result because the mole ratio of AlCl3 to NaHCO3 in the balanced equation is 1:1. Option c, however, would produce a different amount of Al(OH)3 because the mole ratio is different.