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Natali [406]
3 years ago
14

If you started with 35.0 grams of H2S and 40.0 grams of O2, how many grams of S8 would be produced, assuming 95 % yield? 8H2S(g)

+4O2(g)→S8(l)+8H2O(g)
Chemistry
1 answer:
Vesnalui [34]3 years ago
4 0
Molar mass

H2S = 34.0 g/mol
O2 = 31.99 g/mol
S8 = 256.52 g/mol

Identifying excess reagent and the limiting of the reaction :

8 H2S(g) + 4 O2(g) = S8(I) +  8 H2O(g)

8 x 34 g H2S -------->  256. 52 g S8
35.0 g ----------------> ??

35.0 x 256.52 / 8 x 34 =

8978.2 / 272 => 33.00 g of S8  

H2S is  the  limiting reactant
---------------------------------------------

4 x 31.99 g O2 --------------- 256.52 g S8
40.0 g O2 --------------------- ??

40.0 x 256.52 / 4 x 31.99 =

10260.8 / 127.96 = 80.16 g of S8  

O2 is the excess reagent is the excess <span>reagent
</span>
------------------------------------------------------------

H2S is the limiting reactant, one that is fully consumed, it is he who determines the mass of S8 produced 

33.0 g ----------- 100%
?? g ------------- 95 %

95 x 33.00 / 100 => 31.35 g

hope this helps!


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What is the concentration of a potassium iodate solution after you complete the following porcedure? Pipette 10 mL of a 0.31 M p
zubka84 [21]

<u>Given:</u>

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4 0
3 years ago
The mineral enargite is 48.41% cu, 19.02% as, and 32.57% s by mass. what is the empirical formula of enargite?
LuckyWell [14K]
Empirical formula is the simplest ratio of whole numbers of components in a compound. 
Assuming for 100 g of the compound 
                                Cu                                 As                             S
mass                      48.41 g                          19.02 g                      32.57 g
number of moles    48.41 / 63.5 g/mol      19.02 / 75 g/mol        32.57 / 32 g/mol 
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divide by the least number of moles 
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once they are rounded off 
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As - 1
S - 4
therefore empirical formula is Cu₃AsS₄
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