We perform CuO reduction with 5% Hydrogen. How to calculate the theoretical consumption of Hydrogen?
We know
CuO + H2 ----- Cu + H2O
It is theoretically one is to one reaction.
If I am taking 20mg of CuO then mmol of CuO taken is
mmol =(wt/mol.wt) = (20/79.545) = 0.2514mmol.
Therefore Hydrogen consumption also has to be 0.2514 mmol. As Hydrogen is only 5% Pure, it will consume 0.2514*100/5 = 5.028mmol. If I have to calculate Hydrogen consumption (mmol) per gram of CuO, then 5.028mmol of H2/CuO weight = 5.028/20 = 0.2514mmol/g.
Is this correct way of doing the calculation.
Hello Sriram Balachandran ,
Basically is correct. The same amount mol of hydrogen and CuO at 100% efficiency. However, I do not understand you calculation of 0.254/5*100 ... this is not amount of mol for H2. It is for total gas where H2 is 5%.
Important is also that it is calculated with 100% efficiency. So you need slightly more gas for this generally (if trying to estimate how big flow of gas you need. You need slightly more than it is calculated).
0.254*100/5 where 0.254 is the mmol of Hydrogen is required if gas is 100% pure, whereas the gas used is 5%. So this calculation done for purity correction.
Ok. Now it is clear. I think it is fine. However, if you want to use this calculated amount for reduction, it is calculated for 100% efficiency of reaction ... all H2 is consumed. I think you need slightly more of gas because when concentration of H2 drop down too much during reaction the rate of reaction drop down too and some CuO stayed not reduced.
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