How Much Ore is Needed to Extract 47.0kg of Antimony?

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SUMMARY

The discussion centers on calculating the amount of ore needed to extract 47.0 kg of antimony (Sb) from stibnite (Sb2S3), which contains 19.6% of the mineral. The correct calculation involves using the molar mass of Sb2S3, which is approximately 339.73 g/mol, not 153.8 g/mol as initially stated. The final result for the required ore mass is approximately 151.473 kg, but the significance of significant figures must be considered based on the precision of the input data.

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An Ore contains 19.6 %...

Question:

An ore contains 19.6% of the mineral stibnite, Sb2S3, which is a source of the element Sb. how much ore must be processed in order to obtain 47.0kg of Sb?My Work:

(47000g Sb)((153.825g Sb2S3/ 2 (121.76 g Sb)) (100g ore/19.6g Sb2S3)

which equals 151472.5983g
151.473 Kg
i just want to know if my though process is correct and if i am suppose to multiply the mass of Sb becasue it has 2 moles ! please let me know ! thanks ! and if my Sig Fig is correct in the final answer
 
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alicia113 said:
i just want to know if my though process is correct

(47000g Sb)((153.825g Sb2S3/ 2 (121.76 g Sb)) (100g ore/19.6g Sb2S3)

It is not entirely wrong.

if i am suppose to multiply the mass of Sb becasue it has 2 moles !

Yes for 2, but your molar mass of Sb2S3 is completely off. 153.8 would be a molar mass for SbS.

which equals 151472.5983g
151.473 Kg

if my Sig Fig is correct in the final answer

No. How many sig figs in the data?
 


Borek said:
It is not entirely wrong.



Yes for 2, but your molar mass of Sb2S3 is completely off. 153.8 would be a molar mass for SbS.





No. How many sig figs in the data?


oh ok.. so i multiply my sb by 2 and s by 3 .. (their masses and then add them together like normally)
 


alicia113 said:
oh ok.. so i multiply my sb by 2 and s by 3 .. (their masses and then add them together like normally)

That's how you calculate molar mass, yes.
 

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