S8 + 8 O2 --> 8 SO2
2 SO2 + O2 --> 2 SO3
SO3 + H2O --> H2SO4
You find a source of sulfur that is 65.0% sulfur by mass. How many grams of this starting material would be required to produce 87.0 g of sulfuric acid? Explain and show work.
Moles of sulfuric acid required = 87.0g/98.0785g/mol = 0.887045moles
Moles of SO3 required = moles of sulfuric acid = 0.887045
Moles of SO2 required = (2/2)×moles of SO3 = 0.887045
Moles of S8 = (1/8)×moles of SO2 = (1/8)×0.887045
= 0.11088mol
Mass of sulfur required = number of moles × molar mass = 0.11088mol × 256.52g/mol = 28.4431 g
Amount of starting material required = (100/65)×28.4431g
= 43.7586 g
= 43.8 g. (Answer)
Sulfuric acid can be produced by the following sequence of reactions. S8 + 8 O2 --> 8 SO2 ...
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For the reaction represented by the equation SO3+H2O>H2SO4, how many grams pf sulfuric acid can be produced from 320 g of water and 1064 g of sulfuric trioxide
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