Sulfur reacts with oxygen to yield SO3 as shown in the equation below;
2S(g)+ 3O2(g) = 2SO3(g)
From part A 7.49 g of S were used.
The atomic mass of sulfur is 32.06 g/mol
Hence, the number of moles of sulfur used
7.49 / 32.06 = 0.2336 moles
The mole ratio of S : SO3 is 1:1
Thus the mass of SO3 will be ( 1 mol of SO3= 80.06 g)
0.2336 moles × 80.06 = 18.7 g
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Answer:
Avogadro's number or Avogardro’s constant
Explanation:
I’m pretty sure this is correct if it’s not I’m sorry lol.
In chemical reactions, the actual yield is not the same as the expected yield . Actual yield is lower than the theoretical yield . Then we have to find the yield percentage. To see what percentage of the theoretical yield is the actual yield.
Percent yield = actual yield / theoretical yield x 100%
Percent yield = 24.6/55.9 x100%
Percent yield = 44%
