Question

How many gram atoms of sulphur & oxygen are needed to prepare 6.023× 10^24 molecules of So2?​

Answer 1

Answer:

6.022*10^24 molecules of SO2 is approximately 10 mols of SO2. The molar mass of sulfur is 32g/mol and there is 1 mol of sulfur in each mol of SO2. Therefore, the chemist needs 10 mols of sulfure.

Answer 2

Answer:10 gram atom of Sulphur and 320 gram of oxygen

Explanation:

s+o2-------> so2 (balanced equation)

no. of moles of so2= (6.022*10^24)/(6.022*10^23)= 10 mole

1 mole of so2 is obtained by reacting 1 mole of Sulphur with 1 mole of O2

10 mole of so2 is obtained by reacting  10 mole of S with 10 mole of O2

Hence, gram atoms of Sulphur= 10 ( gram atoms means no of moles of S)

now,

mass of Oxygen = no. of moles of o2*molar mass of o2

                           =10*32

                           =320 g

Therefore, the mass of oxygen needed is 320 g.