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
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?
4 answers
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.
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?
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.
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.
1 answer