Question

Determine the limiting reactant and calculate the number of grams of sulfur dioxide, so2, that can be formed when 27.3 g of methane thiol, ch3sh, reacts with 38.6 g of oxygen, o2.

Answer 1

When Methane thiol, CH₃SH, reacts with O₂, 25.8 grams of Sulfur dioxide, SO₂ is formed and limiting reactant is Oxygen, O₂.

The balanced chemical reaction is:

CH₃SH(g) + 3 O₂(g) → 2 H₂O(g) + CO₂(g) + SO₂(g)

             Molecular weights:       Actual/given mass:

CH₃SH = 48.11 g/mol                    27.3 g      

O₂         = 32.01 g/mol                    38.6 g  

H₂O      = 18.01 g/mol            

CO₂      = 44.01 g/mol                  

SO₂       = 64.06 g/mol

Number of moles =    Given mass    

                                  Molecular mass

So,

moles of methane thiol = 27.3 ÷ 48.11= 0.56 moles

moles of oxygen = 32.01÷38.6 = 0.82 moles

From the reaction,

1 mole of CH₃SH reacts with 3 moles of O₂  to give 1 mole of SO₂

Thus, 0.56 moles of CH₃SH reacts to form 1 × 0.56 = 0.56 moles of SO₂

Mass of SO₂ produced is 0.56 × 64.06 g = 35.86 g

moles of SO₂ = 35.86 ÷ 64.06 = 0.55 moles

So,