Someone help me please I need to finish this by today

Option A

Explanation:

First calculate the moles of each reactant,

Moles of SO2 = 544.5 * 1 mole / 64.06 g,

Moles of SO2 = ( About ) 8.5 Moles,

Moles of O2 = 160 * 1 mole / 32 g,

Moles of O2 = 5 Moles,

Now the molar ratio of SO2 to SO3 would be 2 : 2, so that you could say SO3 = ( About ) 8.5 Moles. Respectively, the molar ratio of O2 to SO3 would be 1 : 2, so that SO3 = 5 Moles. The limiting reactant would thus be SO2.

Using that information, we can now calculate the yield of SO3,

Yield of SO3 = 8.5 mole * 80.06 g / 1 mole = 680.5 grams,

Percent Yield = ( 382 / 680.5 ) * 100 = 56.1 %

Solution = Option A

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