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
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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>
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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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Explanation:
170-95.3 = 74.7
that means
74.7% of water in hydrate
Answer:
true ?
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Answer:
Please find the explanation to this question below.
Explanation:
