During the chemical reaction given below 21.71 grams of each reagent were allowed to react. Determine how many grams of the excess reagent are left over after the reaction is complete. 2SO2()+O2()⟶2SO3()

16.32 g of as excess reagent are left.

Explanation:

To calculate the moles :

According to stoichiometry :

2 moles of require = 1 mole of

Thus 0.34 moles of will require=  of

Thus is the limiting reagent as it limits the formation of product and is the excess reagent.

Moles of left = (0.68-0.17) mol = 0.51 mol

Mass of

Thus 16.32 g of as excess reagent are left.

it's all four options.