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!
<span>The activation energy was reached was 10:45 a.m. The additional energy did not affect the reaction.</span>
Answer:
The percent composition is 21% N, 6% H, 24% S and 49% O.
Explanation:
1st) The molar mass of (NH4)2SO4 is 132g/mol, and it represents the 100% of the mass composition.
In 1 mole of (NH4)2SO4, there are:
- 2 moles of N.
- 8 moles of H.
- 1 mole of S.
- 4 moles of O.
2nd) It is necessary to calculate the mass of each element, multiplying its molar mass by the number of moles:
- 2 moles of N (14g/mol) = 28g
- 8 moles of H (1g/mol) = 8g
- 1 mole of S (32g/mol) = 32g
- 4 moles of O (16g/mol) = 64g
3rd) With a mathematical rule of three we can calculate the percent composition of each element in the molecule of (NH4)2SO4:
In this case, we can calculate the percent composition of Oxygen by subtracting the other percentages, since the total must be 100%.
So, the percent composition is 21% N, 6% H, 24% S and 49% O.
<span>The electrons get energy by the potential</span><span> force when we apply potential difference to a conductor moving from low to high therefore, the electrons move one to another creating electricity.</span>
Volatility is a chemical property. The other ones are chemical properties
