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?
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.
To solve this stoichiometry problem, we need to follow a step-by-step approach. Here's how to find the amount of aluminum chloride formed:
Step 1: Write and balance the chemical equation for the reaction:
Aluminum nitrite (Al(NO2)3) + Ammonium chloride (NH4Cl) → Aluminum chloride (AlCl3) + Nitrogen gas (N2) + Water (H2O)
Step 2: Calculate the molar mass of each compound:
Aluminum nitrite (Al(NO2)3): 26.98 g/mol (Al) + 3 * 14.01 g/mol (N) + 6 * 16.00 g/mol (O) = 212.98 g/mol
Ammonium chloride (NH4Cl): 14.01 g/mol (N) + 4 * 1.01 g/mol (H) + 1 * 35.45 g/mol (Cl) = 53.49 g/mol
Step 3: Convert the given masses of aluminum nitrite and ammonium chloride to moles:
Moles of aluminum nitrite = mass / molar mass = 43 g / 212.98 g/mol ≈ 0.202 mol
Moles of ammonium chloride = mass / molar mass = 43 g / 53.49 g/mol ≈ 0.805 mol
Step 4: Determine the limiting reactant:
To find the limiting reactant, compare the mole ratios of the reactants from the balanced equation. The balanced equation tells us that the ratio of moles of aluminum nitrite to moles of ammonium chloride is 1:1. Since the molar ratio is the same as the given ratio, the reactants are in a 1:1 molar ratio, and the limiting reactant is the one with the fewest moles. In this case, aluminum nitrite is the limiting reactant with 0.202 mol.
Step 5: Use the stoichiometry of the equation to find the moles of aluminum chloride:
From the balanced equation, the molar ratio of aluminum nitrite to aluminum chloride is 1:1. Therefore, the moles of aluminum chloride formed will also be 0.202 mol.
Step 6: Convert moles of aluminum chloride to grams:
Mass of aluminum chloride = moles * molar mass = 0.202 mol * (26.98 g/mol + 3 * 35.45 g/mol) ≈ 24.06 g
Therefore, approximately 24.06 grams of aluminum chloride will be formed when 43 grams of aluminum nitrite and 43 grams of ammonium chloride completely react.
To find the amount of aluminum chloride formed, we need to follow a series of steps:
Step 1: Write the balanced chemical equation for the reaction:
2Al(NO3)3 + 6NH4Cl → 2AlCl3 + 6NH3 + 9H2O + N2
Step 2: Calculate the molar mass of each compound involved:
Al(NO3)3: 1 atom of Al (26.98 g/mol) + 3 × (1 atom of N (14.01 g/mol) + 3 atoms of O (16.00 g/mol) = 212.99 g/mol
NH4Cl: 1 atom of N (14.01 g/mol) + 4 atoms of H (1.01 g/mol) + 1 atom of Cl (35.45 g/mol) = 53.49 g/mol
AlCl3: 1 atom of Al (26.98 g/mol) + 3 atoms of Cl (35.45 g/mol) = 133.33 g/mol
Step 3: Determine the limiting reactant:
To find the limiting reactant, compare the moles of aluminum nitrite to the moles of ammonium chloride. Convert the given masses to moles:
Moles of Al(NO3)3 = mass of Al(NO3)3 / molar mass of Al(NO3)3 = 43 g / 212.99 g/mol = 0.2020 mol
Moles of NH4Cl = mass of NH4Cl / molar mass of NH4Cl = 43 g / 53.49 g/mol = 0.8049 mol
The ratio between Al(NO3)3 and NH4Cl in the balanced equation is 2:6, which simplifies to 1:3. Since the ratio of moles of Al(NO3)3 to NH4Cl is less than 1:3, Al(NO3)3 is the limiting reactant.
Step 4: Calculate the moles of the product formed:
From the balanced equation, we know that 2 moles of Al(NO3)3 produce 2 moles of AlCl3.
Therefore, the moles of AlCl3 formed = moles of Al(NO3)3 x (2 moles AlCl3 / 2 moles Al(NO3)3) = 0.2020 mol x (2/2) = 0.2020 mol
Step 5: Convert moles of AlCl3 to grams:
Mass of AlCl3 = moles of AlCl3 x molar mass of AlCl3 = 0.2020 mol x 133.33 g/mol ≈ 26.93 g
Therefore, approximately 26.93 grams of aluminum chloride are present after the reaction is complete.