Question

for each of the following likely the brands of indicated product when 22.0 gram of the first reactant and 39.5 when will the second reactant is used
a) 2SO2(g)+O2(g)--> 2SO3(g)(SO3)
B) 3fe(s)+4H2O(l)-->Fe3O4(s)+4H2(g)(Fe3O4)
c) C7H16(g)+11O2(g)-->7CO2(g)+8H2O(g)(H2O)

Answer 1

Here we hae to find the amount of products produced from the reactants whose mass given in above question.

a)

2SO₂(g)+O₂(g)--> 2SO₃(g)

Mass of SO₂ = 22.0 g

Mass of O₂ = 39.5 g

Let us find the limiting reactant among these two

Moles of SO₃ produced from 22.0 g SO₂ =

= 0.343 mol SO₃

Moles of SO₃ produced from O₂ =

Moles of SO₃ = 2.47 mol SO₃

Limiting reactant is the = SO₂

So mass of SO₃ produced from 22 g SO₂ =

Mass of SO₃ = 27.5 g SO₃

B)

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

Mass of Fe = 22.0 g

Mass of H₂O = 39.5 g

Let us find the limiting reactant among these two

Moles of Fe₃O₄ produced from 22.0 g Fe =

= 0.131 mol Fe₃O₄

Moles of Fe₃O₄ produced from H₂O  =

Moles of Fe₃O₄ = 0.549 mol Fe₃O₄

Limiting reactant is the = Fe

So mass of Fe₃O₄ produced from 22 g Fe =

Mass of Fe₃O₄= 30.3g Fe₃O₄

C)

C₇H₁₆(g) +11O₂(g) ----> 7CO₂(g)+8H₂O(g)

Mass of C₇H₁₆ = 22.0 g

Mass of O₂ = 39.5 g

Let us find the limiting reactant among these two

Moles of H₂Oproduced from 22.0 g C₇H₁₆ =

= 1.76 mol C₇H₁₆

Moles of H₂O produced from O₂ =

Moles of H₂O  = 0.898 mol H₂O

Limiting reactant is the = O₂

So mass of H₂O produced from 39.5 g O₂ =

Mass of H₂O = 16.2 g H₂O