Question

please answer q5 and 7

Answer 1

5) no of mole of atoms = 3 × mole of SO2
= 3 × 0.1 = 0.3

no of atoms = mole of atoms × 6.023 × 10²³ = 0.3 × 6.023 × 10²³ = 1.806 × 10²³

7) 2 Mg + O2 →2 MgO

here we see 2 mole of Mg react with 1 mole of O2

so, 48g of Mg react with 32 g of O2

so, 1 g of Mg react with 32/48 g of O2

e.g 1 gram Mg complete react with 0.667 g of O2
but here given only 0.56g of O2
so, O2 is limiting reagent

now, if 0.56 g complete react then
mass of Mg will require = 48/32× 0.56
= 0.84 g

hence, Mg left in excess
amount of Mg left = 1-0.84 = 0.16 g