Answer:
5.16 gm of SO3 formed with 2 g of S
Explanation:
Mole weight of S in the equation = 2 * 32 = 62 gm
Mole weight os O2 in the equation 6 * 16 =96 gm
From the BALANCED equation the grams of S to O2 is
62 to 96 so 2 g of S will need approx 3 gm of O2
this shows that S is the limiting reactant------>
there will be O left over (approx 1 gram)
SO3 mole weight produced from the equation is 2 (32)(3*16) = 160 gm
62 gm of S produces 160 gm of SO3
62/160 = 2 / x x = 5.16 gm of SO3 are formed
