Question

When a mixture of 10 mole of SO2, 15 mole of O2 was passed over catalyst, 8 mole of SO3 was formed. how many mole of SO2 and O2 did not enter into Combination.

Answer 1

refer the attachment

Answer 2

Answer:

2 moles of SO2 and 11 moles of O₂ did not take part in the reaction

Explanation:

Molar mass of S = 32.06u

Molar mass of O= 16.00u

Molar mass of SO₂ = 64.06u

Mass in grams of 10 moles of SO₂ = 10 x

Mass in grams of 10 moles of SO₂ = 10 x32.06 320.6 g

Mass in grams of 15 moles O₂ = 15 × 32.00

Mass in grams of 15 moles O₂ = 15 × 32.00= 960 g

Mass of S that went in = 10 x 32.06 = 320.6 g

Mass of S that came out in the form of

SO3=8 x 32.06 = = 256.48 g

Mas of S that did not take part in the reaction = 320.6-256.48 64.12 g = 2 moles of S

Now S took part in the reaction in the form of SO2 so if there were 2 moles of S that didn't take part in reaction then there were 2 moles of SO2 that did not take part in the reaction. Now if SO2 is taken out from SO3

then we have only O so the mass of

O₂ that was used in the reaction

8 (No. of moles of O in SO3) × 16.00

(molar mass of oxygen) = 128 g Mass of oxygen that went in = 15 x 32.00 (molar mass of O2) 480.00 g

Therefore the mass of oxygen that was not used was = 480.00-128.00 352 g = 11 moles of O2

Therefore 2 moles of SO2 and 11 moles of O₂ did not take part in the reaction.

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