Stoichiometry

by Tam Le

REACTION : ALUMINUM NITRATE and SODIUM CARBONATE

Type of Reaction : Double Replacement

Al(NO3)3 + Na2(CO3) -> Al2(CO3)3 + NA(NO3)

Double Replacement is the cations and anions of two different compounds switch places

AB + XY ---> AY + XB

The outside portions (the cation A and anion Y) combine to make a formula called AY.

The inside portions (the anion B and the cation X) switch order so that X (postively charged) goes first and B (negatively charged) goes second making a formula called XB

So Al(NO3)3 -> Al2(CO3)3 ; Na2(CO3) -> NA(NO3)

BALANCING, NAMING.

BALANCING:

2Al(NO3)3 + 3Na2(CO3) -> Al2(CO3)3 + 6Na(NO3)

- Al in the left side has 1 mole but on the right side has 2 moles so we have to multiply the Al on the left side by 2.

- (NO3) compound in the left side has 3 moles but on the right side has 1 moles so we have to multiply the (NO3) compound on the right side by 3 to make it balanced

- Na in the left side has 2 moles but on the right side has only 1 mole so we have to multiply the Na on the right side by 2

- (CO3) compound in the left side has 1mole but on the right side has 3 moles so we have to multiply the (CO3) compound on the left side by 3 .

- (Na) now on the left side has 6 moles because we balanced (CO3) before. So we have to multiply Na on the right side by 6. (Na= 2x3=6)

- the (NO3) is now balanced. The whole equation is now balanced

Reference link/ Examples

Balancing Chemical Equations (Double Replacement Reactions)

IUPAC NAME OF THE REACTANTS AND PRODUCTS

Al(NO3): Aluminum nitrate ; Al2(CO3)3: aluminum carbonate

Na2(CO3): Sodium Carbonate ; Na(NO3): Sodium Nitrate.

=> Element + Name of the compound (Nitrate and Carbonate)

MOLAR MASS

Al : 26.982 g/mole

N : 14.007g/mole

O : 15.999g/mole

Na: 22.990g/mole

C : 12.011g/mole


Al(NO3)3 : 26.982+(3*14.007)+(9*15.999) = 212.994g/mole

Na2(CO3): (2*22.990)+12.011+(3*15.999) = 105.988g/mole

Al2(CO3)3: (2*26.982)+(3*12.011)+(9*15.999) = 233.988g/mole

NaNO3 : 22.990+14.007+(3*15.999)=84.994g/mole

WATCH THIS VIDEO TO HAVE A GENERAL IDEA OF MOLE TO MOLE CONVERSIONS STROICHIOMETRY.

Stoichiometry - Moles to Moles (using a balanced equation) | www.whitwellhigh.com

MOLE TO MOLE CONVERSIONS

For example : if 12.31 moles of Aluminum Nitrate are used, how much moles does Sodium Nitrate is needed?
12.31 moles Al(NO3)3 x 6 moles Na(NO3) /
---------------------------------- 2 moles Al(NO3)3 => 36.93 moles NaNO3
Number 6 and 2 is taken from the balanced equation. We use coefficiens as the ratio to keep things equal

MASS TO MASS CONVERSIONS

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LET'S TRY

For example: If you have 12.1gram of Al(NO3)3 , how many grams Na2CO3 could you make?
12.1gAl(NO3)3 x 1 mole Al(NO3)3 x 3 moles Na2(CO3) x 105.988g Na2(CO3) /
-------------------- 212.994g Al(NO3)3 x 2 moles Al(NO3)3 x 1mole Na2(CO3)

- We use the given which is 12.1g of Al(NO3)3 as a tool to find the mass Na2CO3.
- We put it first. Because we're already have the mass so we want to get rid of it. Write the molar mass of Al(NO3) under and diagonal it (212.994g/mole) (We use the molar mass because it relates to the mole which we can transfer to the mole of the compound we wanted to find the mass of it)
- Then we write 1 mole Al(NO3)3 above its molar mass. (212.992 gram in 1 mole of Al(NO3)3)
- Diagonally, we write 2 moles of Al(NO3) which is the ratio
- Above what we just wrote, write 3 moles Na2(CO3). (This is the ratio, like we convert from mole to mole).
- Again, diagonally we write 1 mole of Na2(CO3)
- Above we write 105.988g/mole Na2(CO3) ( because we're finding the mass of Na2(CO3) so we use the molar mass and put it at the top. because all the things we wrote on the bottom will be eliminated, and will be got rid of, we were trying to write the thing that we wanted to get rid of at the bottom.)
-To find the answer, we multiply the things on the top and divided by all the things at the bottom.
So we'll have: 12.1g Al(NO3)3 x 1mole Al(NO3)3 x 3 mole Na2(CO3) x 105.988gNa2(NO3) / [ 212.994gAl(NO3)3 x 2 moles Al(NO3)3 x 1 mole Al2(CO3)3 = 9.03g Na2(CO3).

LIMITING AND EXCESS REACTANTS AND THEORETICAL YIELD.

For example, we have 12.3 Al(NO3)3 and 12.3g Na2(CO3) , how much Al2(CO3)3 can we make.

So we do mass to mass of 12.3g Al(NO3)3 to find the mass of Al2(CO3)3 first. And then we do another conversion from 12.3g Na2(CO3) to the mass of Al2(CO3)3. We need to do it twice to find out which answer is less than another one. If its smaller than another one. The reactant we used to find the mass of Al2(CO3)3 is the limiting and another one is excess.

--12.3g Al(NO3)3 x 1 mole Al(NO3)3 x 1 mole Al2(CO3)3 x 233.988g Al2(CO3)3 / 212.994g Al(NO3)3 x 2 moles Al(NO3)3 x 1mole Al2(CO3)3 = 6.75g Al2(CO3)3

-- 12.3g Na2(CO3) x 1 mole Al(NO3)3 x 1 mole Al2(CO3)3 x 233.988g Al2(CO3)3 / 105.988g Na2(CO3) x 3moles Na2(CO3) x 1mole Al2(CO3)3 = 9.05g Al2(CO3)3


Al(NO3)3 makes a smaller mass than Na2(CO3) so : Al(NO3)3: limiting ; Na2(CO3); excess.

and 6.75g Al2(CO3) is the theoretical yield because we can only make 6.75g with 12.3g Al(NO3)3. So that is the number we're supposed to make.

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PERCENT YIELD

% YIELD = (ACTUAL YIELD / THEORETICAL YIELD) x 100

so lets say we actually made 8.02g of Al2(CO3)3 in the experiment. So the percent yield of Al2(CO3)3 is = (8.02/6.75)x100= 118.81%.

The number is higher than 100 because we made more than the mass we're supposed to make.

THE END !! HOPE THIS WILL HELP YOU UNDERSTAND MORE ABOUT STOICHMETRY