ALUMINUM NITRATE & SODIUM CARBONATE
BY- YORDANOS KIFLE / 7TH PERIOD
TYPE OF REACTION
You can think of the reaction as swapping the cations or the anions (but not swapping both, since you would end up with the same substances you started with).
Here is an example of a double replacement reaction:
2Al(NO3)3 (aq) + 3Na2(CO3)(aq)------------->Al2(CO3) (s)( + 6Na(NO3) (aq)
(Aluminum Nitrate + Sodium Carbonate---->Aluminum Carbonate +Sodium Nitrate)
Identifying Double -Replacement reactions is usually fairly straightforward once you can recognize the pattern. Predicting whether the reaction will occur can be trickier, and it helps to be able to recognize some common types of double replacement reactions. Precipitation reactions is one of the types of double replacement reactions. Precipitation reactions produce an insoluble product from two aqueous reactants, and you can identify a precipitation reaction using solubility rules.
HOW TO FIND MOLAR MASS
1. Find the atomic Masses of individual Elements in the Periodic Table
2. Count how many atoms there are for each element
3. Multiply the atomic mass (from the PERIODIC TABLE ) of each element by the number of each element by the number of atoms of that element present in the compound
4.Add it all together and put units of Grams/Mole after the number.
Example Find the Molar Mass of Aluminum Nitrate , Al(NO3)3
Al 1 x 26.982 = 26.982
N 3 x 14.007 = 42.021
O 9 X 15.999 = 143.991
MOLAR MASS = 2212.994 g/Moles
After calculating the molar mass of each compound you should end up with an answer like this------->
Al(NO3)3 = 26.982 g/moles
Na2(CO3) = 105.988 g/moles
Al(CO3)3 = 233.988 g/moles
Na(NO3) = 84.994 g/moles
MOLE TO MOLE CONVERSIONS
In this type of problem, the amount of one substance is given in moles. From this, you are to determine the mass of another substance that will either react with or be produced from the given substance.
Moles of given→Moles of unknown→Mass of unknown
INFORMATION NEEDED TO PERFORM CALCULATION
- Moles and identity of given substance
- Identity of desired substance
- Mole ratio between given substance and desired substance. Remember, mole ratio comes from the balanced chemical equation!
Once we have this information, we multiply the moles of our given by our mole ratio between our desired substance and given substance. This will give us the number of moles of our desired substance:
(Moles of given substance/1) * (Moles of desired substance/moles of given substance) = Moles of desired substance
Q- How many moles of Na(NO3) can be made from 2.2 moles of Al(NO3)3?
A- (2.2 moles Al(NO3)3) * ( 6 Moles Na(NO3) / 2 Moles Al(NO3)3 ) = 6.6 Moles Na (NO3)
MASS TO MASS CONVERSTIONS
- Write a balanced equation for the reaction.
- Write the given mass on a factor-label form.
- Convert mass of reactant to moles of reactant.
- Convert moles of reactant to moles of product.
- Convert moles of product to grams of product.
- Pick up the calculator and do the math
If you have 12.1g of Al(NO3)3, how many grams Na2(CO3) can be produced?
If you followed the steps above you should end up with ------> 9.03g Na2(CO3)
Take a look at another example in the video!
LIMITING AND EXCESS REACTANT
Excess Reactant - The reactant in a chemical reaction that remains when a reaction stops when the limiting reactant is completely consumed. The excess reactant remains because there is nothing with which it can react.
*Do Mass to Mass conversion to figure out which reactant is the limiting and excess.
12.3 g Al(NO3)3-------PRODUCES---------->14.72g Na(NO3)
12.3 g Na2(CO3)------PRODUCES----------> 19.73g Na(NO3)
In that case, Al(NO3)3 is the limiting reactant because it produced the least amount of Sodium Nitrate ; Na2(CO3) is the excess reactant.
The mass of product that you calculate using stoichiometry ; it's what you are suppose to be able to get from reaction(theoretically).
From the example given above the Theoretical Yield would be 14.72g Na(NO3).
WHAT IS PERCENT YIELD?
- The percentage yield is the ratio between the actual yield and the theoretical yield multiplied by 100%. It indicates the percent of theoretical yield that was obtained from the final product in an experiment.
- The percentage yield can be calculated using the mass of the actual product obtained and the theoretical mass of the product calculated using the balanced equation of the reaction
ACTUAL YIELD = 16.93g Na(NO3)
HEORETICAL YIELD= 14.72 g Na(NO3)
Perecent YIeld= (16.93/14.72) x 100= 115.01%
WATCH THIS VIDEO FOR MORE PRACTICE.......