4NH3 + 5O2 = 4NO + 6H2. The gas-phase reaction between ammonia and oxygen occurs only at high temperature. It is referred to as “burning” ammonia, even though no carbon is present. The reaction is carried out in a sealed flask initially containing 88.5 grams of ammonia and 88.5 grams of oxygen.
f. What mass of oxygen remains unreacted?
g. Analysis shows that at total mass of 103.5 grams of products (water and nitrogen monoxide combined) was formed? What is the percent yield for the reaction?
h. What mass (g) of nitrogen monoxide was actually produced?
A quick do-over. I se the question is the same but the parts are different. Please explain to me why you can't the answers I gave you and use them to solve a similar problem. Asking for NO instead of H2O shouldn't be a problem. I need to understand what you're having trouble understanding.
I answered this for you an hour after you posted it the first time. Here is a url to get to it.
https://www.jiskha.com/questions/1862603/i-also-need-help-with-this-please-the-gas-phase-reaction-between-ammonia-and-oxygen-occurs
Ok I understood the rest. But I'm confused with number h.
This is what I got
66.5 g NO * (82/100) = 54.5g
I got that none of nitrogen monoxide was produced.
Dr Bob is this correct?
To do h we need the percent yield from g and I don't remember the numbers from the earlier part so I'll need to repeat some of the earlier part. You made a typo in the equation so I've redone it here.
4NH3 + 5O2 = 4NO + 6H2O
g. I remember O2 is the limiting reagent. If O2 is the limiting reagent that means all of it is used and that means none is left over. We had 88.5 g O2 which is 88.5/32 = 2,76 mols O2. That will produce how much NO and how much H2O.
mols NO produced theoretically = 2.76 x (4 mols NO/5 mols O2) = 2.76 x 4/5 = 2.21
mols H2O produced theoreticlly = 2.76 x (6 mols H2O/5 mols O2) = 2.76 x 6/5 = 3.31
Convert these to grams.
grams NO = mols NO x molar mass NO = 2.21 x 30 = 66.3 g NO
grams H2O = 3.31 mols H2O x 18 = 59.6
Total grams NO + grams H2O = 125.9. The is the theoretical yield.
Total grams NO + grams H2O from the problem which is the actual yield. That is actual yield of 103.5 g
%yield = (actual yield/theor yield)*100 = (103.5/125.9)*100 = 82.2%.
h. I think you did it right but you have the wrong conclusion. My calculations are as follows:
%yield = (actual yield/theore yield)*100
82.2 = (AY/TY)*100
0.822 = (AY/TY)
0.822*TY = AY = 0.822*66.3 = 54.5 g NO actually produced.
Your calculations show that 54.5 g NO was actually produced. The theoretical yield was 66.3 g but you only produced 54.5 g since the reaction was only 82.2% effective. Don't know why you thought none was produced.
I hope this is clear to you. Let me know, in detail, if it isn't clear. Good luck in your chem class, It's a great course and a great career.
To answer these questions, we need to use stoichiometry and the concept of percent yield.
f. What mass of oxygen remains unreacted?
To determine the mass of oxygen that remains unreacted, we first need to calculate the amount of ammonia and oxygen used in the reaction.
Molar mass of NH3 (ammonia) = 17.03 g/mol
Molar mass of O2 (oxygen) = 32.00 g/mol
We are given that the reaction is carried out with 88.5 grams of both ammonia and oxygen.
Step 1: Calculate the moles of ammonia and oxygen.
Moles of NH3 = Mass of NH3 / Molar mass of NH3
Moles of NH3 = 88.5 g / 17.03 g/mol ≈ 5.19 mol
Moles of O2 = Mass of O2 / Molar mass of O2
Moles of O2 = 88.5 g / 32.00 g/mol ≈ 2.77 mol
Step 2: Use the balanced chemical equation to determine the mole ratio between ammonia and oxygen.
The balanced equation is: 4NH3 + 5O2 -> 4NO + 6H2
From the equation, we can see that the ratio of NH3 to O2 is 4:5.
Step 3: Determine which reactant is the limiting reagent.
Comparing the mole ratio (5:4) with the moles calculated in step 1, we can see that ammonia is in excess. Therefore, ammonia is the limiting reagent, and oxygen is the reactant that remains unreacted.
Step 4: Calculate the moles of oxygen used in the reaction.
Moles of O2 used = Moles of NH3 * (5/4)
Moles of O2 used = 5.19 mol * (5/4) ≈ 6.49 mol
Step 5: Calculate the mass of oxygen that remains unreacted.
Mass of O2 remaining = (Total moles of O2 - Moles of O2 used) * Molar mass of O2
Mass of O2 remaining = (2.77 mol - 6.49 mol) * 32.00 g/mol ≈ -138.24 g
Since we obtained a negative mass, it indicates that all the oxygen was consumed in the reaction, and there is no oxygen remaining unreacted.
g. What is the percent yield for the reaction if a total mass of 103.5 grams of products (water and nitrogen monoxide combined) was formed?
The percent yield can be calculated using the following formula:
Percent yield = (Actual yield / Theoretical yield) * 100
Step 1: Calculate the theoretical yield of the nitrogen monoxide (NO) using the balanced equation.
From the balanced equation, the mole ratio of NH3 to NO is 4:4. Therefore, the number of moles of NO produced is equal to the number of moles of NH3 used.
Moles of NO produced = Moles of NH3 used
Moles of NO produced = 5.19 mol
Step 2: Calculate the molar mass of NO.
Molar mass of NO = 30.01 g/mol
Step 3: Calculate the theoretical yield of NO in grams.
Theoretical yield of NO = Moles of NO produced * Molar mass of NO
Theoretical yield of NO = 5.19 mol * 30.01 g/mol ≈ 155.82 g
Step 4: Calculate the percent yield.
Percent yield = (Actual yield / Theoretical yield) * 100
Percent yield = (103.5 g / 155.82 g) * 100 ≈ 66.50%
Therefore, the percent yield for the reaction is approximately 66.50%.
h. What mass (g) of nitrogen monoxide was actually produced?
The actual yield is given as 103.5 grams, so the mass of nitrogen monoxide actually produced is 103.5 grams.