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Sagot :

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

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