Question

For each of the following reactions, calculate the grams of indicated product when 25.0 g of the first reactant and 41.0 g of the second reactant are used.

a.) 2SO₂ (g) + O₂ (g)------> 2SO₃ (g) (SO₃)

m(SO₃)=__________

b.) 3Fe (s) + 4H₂O (l) -------> Fe₃O₄ (s) + 4H₂ (g) (Fe₃O₄₎

m(Fe₃O₄)=__________

Answer 1

a.) 2SO₂ (g) + O₂ (g) 2SO₃ (g)

MOlar mass of SO₂ = 32 + (2x16) = 64 g/mol

MOlar mass of O₂ = 2x16 = 32 g/mol

MOlar mass of SO₃ = 32 + (3x16) = 80 g/mol

According to the balanced equation ,

2 moles of SO₂ reacts with 1 mole of O₂

                                    OR

2 x64 (= 128) g of SO₂ reacts with 1x32 (=32) g of O₂

25.0 g of SO₂ reacts with M g of O₂

M = ( 32x25.0 ) / 128

   = 6.25 g of O2

So 41.0 - 6.25 g of O₂ left unreacted So O₂ is the excess reactant

Since all the mass of SO₂ reacted it is the limiting reactant.

From the balanced equation ,

2 moles of SO₂ produces 2 moles of SO₃

                                OR

2x64 g of SO₂ produces 2x80 g of SO₃

25.0 g of SO₂ produces N g of SO₃

N = (2x80x25.0 ) / (2x64)

   = 31.25 g of SO₃

(b) 3Fe (s) + 4H₂O (l) Fe₃O₄ (s) + 4H₂ (g)

MOlar mass of Fe = 55.8 g/mol

MOlar mass of H₂O = (2x1) + 16 = 18 g/mol

MOlar mass of Fe₃O₄ = (3x55.8) + (4x16) = 231.4 g/mol

According to the balanced equation ,

3 moles of Fe reacts with 4 mole of H₂O

                                    OR

3x55.8(=167.4) g of Fe reacts with 4x18 (=72) g of H₂O

25.0 g of Fe reacts with M g of H₂O

M = ( 72 x 25.0 ) / 167.4

   = 10.75 g of H₂O

So 41.0 - 10.75 g of H₂O left unreacted So H₂O is the excess reactant

Since all the mass of Fe reacted it is the limiting reactant.

From the balanced equation ,

3 moles of Fe produces 1 mole of Fe₃O₄

                                OR

3x55.8 g of Fe produces 1x231.4 g of Fe₃O₄

25.0 g of Fe produces N g of Fe₃O₄

N = (1x231.4x25.0 ) / (3x55.8)

   = 34.5 g of Fe₃O₄