Posted by Jamie on Thursday, May 13, 2010 at 9:27pm.
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.
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