Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: mdawg467 on September 12, 2011, 04:23:14 AM
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Hi guys,
Here is the question:
The decomposition of sodium chlorate yields oxygen gas and sodium chloride. If the yield is 85% at STP, how many grams of sodium chlorate are needed to produce 8.00L of oxygen?
My solution:
2NaClO3 :rarrow: 3O2 +2NaCL
(8.00L O2)(1 mol O2/22.4L)(2 mol NaClO3/3 mol O2)(106.4g/1 mol NaClO3)= 25.3 g NaClO3
My concern:
I feel as I completed this problem correctly, however the fact that I did not incorporate the % yield is making me wondering if I missed a step..any input?
Thanks!
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Hi guys,
Here is the question:
The decomposition of sodium chlorate yields oxygen gas and sodium chloride. If the yield is 85% at STP, how many grams of sodium chlorate are needed to produce 8.00L of oxygen?
My solution:
2NaClO3 :rarrow: 3O2 +2NaCL
(8.00L O2)(1 mol O2/22.4L)(2 mol NaClO3/3 mol O2)(106.4g/1 mol NaClO3)= 25.3 g NaClO3
My concern:
I feel as I completed this problem correctly, however the fact that I did not incorporate the % yield is making me wondering if I missed a step..any input?
Thanks!
Almost there. What does it mean that the reaction only has a 85% yield? If it only had a 50% yield, for instance you can imagine that half your reactants are being wasted, so how much more would you need to get a given amount of material out?
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Almost there. What does it mean that the reaction only has a 85% yield? If it only had a 50% yield, for instance you can imagine that half your reactants are being wasted, so how much more would you need to get a given amount of material out?
Hi sjb,
Thank you for the response.
Since the reaction yields only 85% of the final product we can say that 15% of the reactants are being "wasted" correct?
So in order to get the the 8.00L of O2 from the sodium chlorate I would need to add an additional 15% to my final answer of 25.3g of NaClO3..so my final answer should be 29.1g of NaClO3.
Is this the correct way of approaching it?
Thanks again.
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Since the reaction yields only 85% of the final product we can say that 15% of the reactants are being "wasted" correct?
So in order to get the the 8.00L of O2 from the sodium chlorate I would need to add an additional 15% to my final answer of 25.3g of NaClO3..so my final answer should be 29.1g of NaClO3.
Close, but wrong. Assuming you use 29.1g and 85% will react, that means mass of the reagent use is 0.85*29.1g=24.7g, not 25.3g.
You don't add 15%, you know 25.3g is 85% and you want to know what is 100%. That's not the same thing. But it is not difficult, just think for a moment how to calculate 100% if you know 85%. Hint: you can use ratios for that.
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Close, but wrong. Assuming you use 29.1g and 85% will react, that means mass of the reagent use is 0.85*29.1g=24.7g, not 25.3g.
You don't add 15%, you know 25.3g is 85% and you want to know what is 100%. That's not the same thing. But it is not difficult, just think for a moment how to calculate 100% if you know 85%. Hint: you can use ratios for that.
Hi Borek,
Thanks for the reply.
Okay I think I understand it now!
So if I set it up like a ratio as you suggested I come up with the following: (25.3/X)=(85/100), and once solved I get X=29.8g. This checks out to be correct as 29.8x0.85=25.3g!
So I would need a total mass of 29.8 grams of NaClO3 to have 8.00L of O2 at these conditions.
Thanks for the help everyone..this forum is full of nice and helpful individuals!