# CHEMISTRY

posted by .

This is STOICHIOMETRY. Please i need help in the worst way. I don't understand this at all and it is due tomorrow, please i need someones help who really understands this.

#1.aluminum nitrite and ammonium chloride react to form aluminum chloride, nitrogen gas, and water. What amount of aluminum chloride is present after 43g of aluminum nitrite reacts with 43g of ammonium chloride have completely?

• CHEMISTRY -

This is a fake limiting reagent problem. It is not a simple stoichiometry problem. However, they TELL you that Al(NO2)3 is the limiting reagent by asking for the amount of NH4Cl remaining after all of the Al(NO2)3 has reacted. [I guess it's a fake limiting reagent problem.]

Step 1. Use the word equation to write a chemical equation, then balance it.
Al(NO2)3 + NH4Cl ==> AlCl3 + N2 + H2O

Then balance it. These are done, usually, by trial and error.
Rule 1 in balancing equations. You may NOT change the subscripts to balance. Once the equation is written, the subscripts are in stone.
Rule 2. You balance ONLY by changing coefficients.
I look and immediately I see 3Cl on the right. I can fix Cl on the left by placing a 3 as a coefficient for NH4Cl.
Al(NO2)3 + 3NH4Cl ==> AlCl3 + N2 + H2O.

Al looks ok. (I usually do H and O last). I have 6 N on the left (3 from Al(NO2)3 and 3 from NH4Cl. So I place a 3 as a coefficient for N2 on the right.
Al(NO2)3 + 3NH4Cl ==> AlCl3 + 3N2 + H2O

Now we tackle H. I see 12 H on the left (3 x 4 = 12). That gets a 6H2O on the right.
Al(NO2)3 + 3NH4Cl ==> AlCl3 + 3N2 + 6H2O

That should do it but I ALWAYS check it.
1 Al both sides.
6 N left and 6 right.
6 O left and right.
12 H left and right.
3Cl left and right. I will rewrite it in bold so we can refer to it later.
Al(NO2)3 + 3NH4Cl ==> AlCl3 + 3N2 + 6H2O

Step 2a. Chemicals react by moles so convert 43 g of what you have [in this case Al(NO2)3] to moles. moles = grams/molar mass. 43/165 = about 0.25 (You do it more accurately.)
[You may ask what we do with the 43 g NH4Cl but since this is a fake limiting reagent problem we leave it as is.]

Step 3. Using the coefficients in the balanced equation, convert moles of what you have (in this case Al(NO2)3 to moles of what you want (in this case moles NH4Cl).
0.25 mol Al(NO2)3 x (3 moles NH4Cl/(1 mole Al2(NO2)3) = 0.25 x (3/1) = 0.75 moles NH4Cl(again you do it more accurately).

Step 4. Now convert moles NH4Cl to grams. g = moles x molar mass.
0.75 moles x (53.5 g NH4Cl/mols NH4Cl) = 42 g
This is the end of the regular stoichiometry problem.

Step 5. The problem asks how much NH4Cl remains. The answer is 43 initially-42 used = 1 g NH4Cl remains un-reacted.

## Similar Questions

1. ### HELP!!!!EMERGENCY WITH THE POEM!!!

do anyone knows a poem named "Instruction for Joining a New Society", which about the desire to become an American?
2. ### Chemistry

Chemical formula for Mercury(II)Chloride, Potassium nitrite, Aluminum hydroxide, Sulfure Dioxide. Let me see if I am right for the ones I completed, okay for Potassium nitrite I put KNO^- and subscript 2 and for Aluminu hydroxide I …

i need to know who the antagonist is in the story money hungry and i need to know what the conflict in the ook is please help me....i really need help i gotta know this by tomorrow!
4. ### Stoichiometry

I am taking an online chemistry class and I do not at all understand how on earth how to do stoichiometry. I just don't get it, its not that I don't know how to do the math, but figuring out how to ger the information. PLEASE HELP!! …
5. ### Chemistry [Science]

Oh gosh... I need help, with like everything... [Show all work] Aluminum nitrite & ammonium chloride react to form aluminum chloride, nitrogen gas and water. What amount of aluminum chloride is present after 43g of ammonium chloride …
6. ### chemistry

Aluminum nitrate and ammonium chloride react to form alluminum chloride, nitrogen, and water. what mass of aluminum chloride is present after 43.0 g of aluminum nitrate and 43.0 g ammonium chloride have reacted completely?
7. ### Chemistry

Aluminum and chlorine gas react together to make Aluminum chloride. What is the limiting reactant if 20.0g of Al and 30.0g of Cl2 are used?