An Ore contains 19.6 %

  • Thread starter alicia113
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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
 

Answers and Replies

  • #2
Borek
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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?
 
  • #3
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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)
 
  • #4
Borek
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28,881
3,429


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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