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....
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I'm sorry...I'm not using his name..I know his one of the best tutors, his help me a lot in understanding chemistry..I wrote like that so he could read my question..and check my work...sorry Ms. Sue...all i was trying to do is to direct the question for Drbob22...sorry once again
To determine the limiting reagent and the theoretical yield of nitrogen gas (N2) in the given chemical equation, you need to follow these steps:
Step 1: Convert the given masses of ammonia (NH3) and copper(II) oxide (CuO) to moles.
- Mass of NH3 = 18.1 g
- Molar mass of NH3 = 17.031 g/mol
- Moles of NH3 = Mass of NH3 / Molar mass of NH3
= 18.1 g / 17.031 g/mol
≈ 1.063 mol NH3
- Mass of CuO = 90.4 g
- Molar mass of CuO = 79.545 g/mol
- Moles of CuO = Mass of CuO / Molar mass of CuO
= 90.4 g / 79.545 g/mol
≈ 1.138 mol CuO
Step 2: Determine the stoichiometric ratio between NH3 and N2.
From the balanced equation: 2 NH3(g) + 3 CuO(s) = N2(g) + 3 Cu(s) + 3 H2O(g)
You can see that 2 moles of NH3 will produce 1 mole of N2.
Step 3: Calculate the theoretical yield of N2 using the limiting reagent concept.
To determine the limiting reagent, compare the number of moles of each reactant with their stoichiometric ratio.
Using NH3:
- Moles of N2 from NH3 = 1.063 mol NH3 * (1 mol N2 / 2 mol NH3)
≈ 0.532 mol N2
Using CuO:
- Moles of N2 from CuO = 1.138 mol CuO * (1 mol N2 / 3 mol CuO)
≈ 0.379 mol N2
Since the moles of N2 from NH3 (0.532 mol) are greater than the moles of N2 from CuO (0.379 mol), NH3 is the limiting reagent. This means that CuO is in excess.
Step 4: Calculate the theoretical yield of N2 using the limiting reagent.
The molar mass of N2 is 28.02 g/mol.
Theoretical yield of N2 = Moles of N2 from the limiting reagent * Molar mass of N2
= 0.532 mol N2 * 28.02 g/mol
≈ 14.94 g N2
Therefore, the correct theoretical yield of N2 is approximately 14.94 g, not 59.68 g as mentioned.
Step 5: Calculate the percent yield of N2.
Given: Actual yield of N2 = 8.52 g (as mentioned)
Percent yield of N2 = (Actual yield / Theoretical yield) * 100
= (8.52 g / 14.94 g) * 100
≈ 57.06%
So, the correct percent yield of nitrogen gas is approximately 57.06%, not 14.28% as mentioned.
Please make sure to double-check your calculations to ensure accuracy.