# ALUMINUM NITRATE & SODIUM CARBONATE

## TYPE OF REACTION

Aluminum Nitrate and Sodium Carbonate is a Double- Replacement Reaction. Double replacement reactions occurs when parts of two ionic compounds are exchanged, making two new compounds. The overall pattern of a double replacement reaction looks like this:*Look at the Figure above*

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

STEPS

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------->

REACTANT

Al(NO3)3 = 26.982 g/moles

Na2(CO3) = 105.988 g/moles

PRODUCT

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

EXAMPLE-

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

The most common type of stoichiometry calculation is a mass-mass problem. Generally, a mass-mass problem looks like this: "given this amount of reactant, how much product will form?"

1. Write a balanced equation for the reaction.
2. Write the given mass on a factor-label form.
3. Convert mass of reactant to moles of reactant.
4. Convert moles of reactant to moles of product.
5. Convert moles of product to grams of product.
6. Pick up the calculator and do the math

QUESTION

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!

Mole-to-mole and Mass-to-mass Conversions

## LIMITING AND EXCESS REACTANT

Limiting Reactant - The reactant in a chemical reaction that limits the amount of product that can be formed. The reaction will stop when all of the limiting reactant is consumed.

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.

Example

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.

STOICHIOMETRY - Limiting Reactant & Excess Reactant Stoichiometry & Moles

## THEORETICAL YIELD

WHAT IS THEORETICAL YIELD?

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
EXAMPLE-

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.......

Percent Yield - Brightstorm Chemistry