Some math help

[Wonch]

Question:

An ore contains 16.0 % of the mineral bismuthinite, Bi2S3, which is a source of the element Bi. How much ore must be processed in order to obtain 12.0 kg of Bi?


I know this is more so a mathematical question, and this is probably why I can't seem to figure out where to start with this question. Any assistance would be awesome. :)

[Òåíü]

An ore contains 16.0 % of the mineral bismuthinite, Bi2S3, which is a source of the element Bi. How much ore must be processed in order to obtain 12.0 kg of Bi?

12 kg - 209 g/mol
x kg - 514 g/mol

х = 29,5 kg

29,5 kg - 16%
y- 100%

y = 184 kg

[Wonch]

12 kg - 209 g/mol
x kg - 514 g/mol

х = 29,5 kg

29,5 kg - 16%
y- 100%

y = 184 kg
I'm sorry but I don't understand your calculations, can you explain further?

[Òåíü]

I don't speak English.
At first we have found weight Bi2S3 from which it is possible to receive 12 kg. Bi.
12 kg - 209 g/mol
x kg - 514 g/mol

х = 29,5 kg

Here 209 g/mol = M(Bi), 514 g/mol = M(Bi2S3).

Now we find weight of ore:

29,5 kg - 16%
y- 100%



y = 184 kg

[jimmyp]

I have a similar question, what is the final answer you're getting here??

[Russ]

Question:

An ore contains 16.0 % of the mineral bismuthinite, Bi2S3, which is a source of the element Bi. How much ore must be processed in order to obtain 12.0 kg of Bi?


I know this is more so a mathematical question, and this is probably why I can't seem to figure out where to start with this question. Any assistance would be awesome. :)

If you assume 100% extraction from the ore, then to get 12kg of Bi we would need 75kg of Bi2S3 (that's pure maths as a ratio). It also assumes the 16% is by mass.

Otherwise, you can calcuate the number of moles of Bi in 12kg, the number of moles of Bi2S3 this will form (given excess S and complete reaction) and the subsequent mass?

[Wonch]

Okay I am still a bit confused by your reply, I can't seem to follow your logic. What is the FINAL answer you're getting here?



I figured in 100g of ore there's 16.0% (this is where I'm confused!!) and found the molar mass of Bi2S3 to be :

209.0 + 209.0 + 32.07 + 32.07 + 32.07 = 514.21 g / mole and therefore I say 1 mole contains 418 g of Bi (or is it just 209.0g?) and I set up my equation as such....


514.21 g / mol Bi2S3 * 100g ore / 16.0g Bi2S3

= 51421 / 16.0
= 3213.8125


.... Is this correct? I think I did something wrong, if I did could you show me the correct method?

Thanks for the reply regardless though!

[Òåíü]

have a similar question, what is the final answer you're getting here??
184 kg

514.21 g / mol Bi2S3 * 100g ore / 16.0g Bi2S3

= 51421 / 16.0
= 3213.8125


.... Is this correct?

Wrong.

[JGK]

Question:

An ore contains 16.0 % of the mineral bismuthinite, Bi2S3, which is a source of the element Bi. How much ore must be processed in order to obtain 12.0 kg of Bi?


MW of bismuthinite = 514

Bismuth (2 Atoms/molecule) accounts for (209 * 2) *100/514 = 81.3% of the total weight of bismuthinite

So if 12 kg of bismuth is refined, this equates to 12 /0.813 =14.8 kg of bismuthinite.

As the original ore contained 16% bismuthinite, 14.8/0.16 = 92.5 kg of the original ore would be required to provide the 12 kg of Bismuth.

[bjhopper]

Hello all,
The answer is in the prior posts but lets clarify it.

Atomic weight of bismuth sulfide is 514.15 ( two Bi and three S )
Atomic weight of Bismuth is 209 so two Bi =418
therefore Bi is 418/514.15 = .813 fraction of pure compound
For 12 kg of Bi you need 12/.813= 14.8 kg of pure compound
14.8 kg /.16 = 92.5 kg of ore

lets check 92.5 x .16 x .81= 12 kg


bjh