consider the following balance chemical equation:

2NH3(g)+3CuO(s)=N2(g)+3Cu(s)+3H2O(g)

if 18.1grams of ammonia reacts with 90.4 grams of copper(II) oxide, which is the limitting reagent?? what is the theorectical yield of nitrogen gasd? if 8.52 g of N2 are formed, what is the % yield of nitrogn gas??

please check if is correct....

limitting reagent is NH3 with 2.13 mol N2
theoretical yield nitrogen gas is 59.68g N2
percent yield of nitrogen gas is 14.28%
please i need to know what i'm doing wrong..thank you....

No, it isn't correct. From the few numbers you show, I have been able to guess that you have the moles right but you used the wrong conversion to convert to moles N2.

moles NH3 = 18.1/17 = about 1.06
moles N2 = 1.06 x (1 mole N2/2 moles NH3) = 1.06/2 = 0.53 moles N2.

moles CuO = 90.4/79.545 = 1.136
moles N2 formed = 1.136 x (1 mole N2/3 moles CuO) = 1.136*(1/3) = 0.379 moles N2.

I will leave it at that and let you finish but it is important for you to understand that you should have divided by 2 and 3 instead of multiplying by 2 and 3 (at least that's what I think you did).

ok..I see what i did wrong...thank u..ur the best!!!

To determine the limiting reagent, we need to compare the moles of ammonia and copper(II) oxide with their respective stoichiometric coefficients.

1. Convert the given masses of ammonia (NH3) and copper(II) oxide (CuO) to moles:

NH3:
18.1 g NH3 * (1 mol NH3 / 17.03 g NH3) = 1.062 mol NH3

CuO:
90.4 g CuO * (1 mol CuO / 79.55 g CuO) = 1.136 mol CuO

2. Calculate the mole ratio between the limiting reagent (NH3) and the product (N2):

From the balanced equation: 2 mol NH3 : 1 mol N2

3. Find the moles of N2 that can be formed from the limiting reagent:

1.062 mol NH3 * (1 mol N2 / 2 mol NH3) = 0.531 mol N2

4. Compare the moles of N2 that can be formed from the limiting reagent (NH3) with the given yield (8.52 g N2):

8.52 g N2 * (1 mol N2 / 28.0134 g N2) = 0.303 mol N2

Since 0.531 mol N2 can be formed from NH3, which is greater than the calculated yield of 0.303 mol N2, NH3 is not the limiting reagent.

5. Calculate the theoretical yield of N2 from the actual limiting reagent. Since CuO is the only reactant remaining, we can determine the theoretical yield directly from its moles:

1.136 mol CuO * (1 mol N2 / 3 mol CuO) * (28.0134 g N2 / 1 mol N2) = 10.54 g N2

Therefore, the theoretical yield of N2 gas is 10.54 grams.

6. Calculate the percent yield of N2 gas:

Percent yield = (Actual yield / Theoretical yield) * 100
Percent yield = (8.52 g N2 / 10.54 g N2) * 100 = 80.88%

So, the correct answer is:

The limiting reagent is CuO (copper(II) oxide).
The theoretical yield of N2 gas is 10.54 grams.
The percent yield of N2 gas is 80.88%.

To determine the limiting reagent in a chemical reaction, you need to compare the number of moles of each reactant to their stoichiometric coefficients in the balanced equation.

Let's calculate the number of moles of NH3 and CuO using their given masses:
Mass of NH3 = 18.1 g
Molar mass of NH3 = 17.03 g/mol
Number of moles of NH3 = 18.1 g / 17.03 g/mol = 1.063 mol

Mass of CuO = 90.4 g
Molar mass of CuO = 79.55 g/mol
Number of moles of CuO = 90.4 g / 79.55 g/mol = 1.136 mol

Now, let's compare the moles of NH3 and CuO to their stoichiometric coefficients in the balanced equation:
According to the balanced equation:
2 moles of NH3 react with 3 moles of CuO to produce 1 mole of N2.

For NH3:
1.063 mol NH3 * (1 mol N2 / 2 mol NH3) = 0.532 mol N2

For CuO:
1.136 mol CuO * (1 mol N2 / 3 mol CuO) = 0.379 mol N2

From the calculations, we can see that NH3 produces more moles of N2 (0.532 mol) compared to CuO (0.379 mol). Therefore, NH3 is the limiting reagent in this reaction.

Now let's calculate the theoretical yield of N2, assuming NH3 is the limiting reagent:
Molar mass of N2 = 28.02 g/mol
Theoretical yield of N2 = 0.532 mol N2 * (28.02 g/mol) = 14.91 g N2

From your calculation, you obtained 59.68 g for the theoretical yield of N2. This value seems incorrect. Double-check your calculations.

To calculate the percentage yield of N2, you need the actual yield of N2 and the theoretical yield.

Given: Actual yield of N2 = 8.52 g

Percentage yield = (Actual yield / Theoretical yield) * 100
Percentage yield = (8.52 g / 14.91 g) * 100 = 57.11%

So, the corrected answer is:
The limiting reagent is NH3.
The theoretical yield of N2 is 14.91 g.
The percent yield of N2 is 57.11%.