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Part II: Limiting Reactant1. Consider the reaction: 5C+2SO2 → CS₂ + 4COIa.) If you had 10 mol of Carbon, how many moles of carbon monoxide would be produced?b.) If you had 10 mol of sulfur dioxide, how many moles of carbon monoxide would beproduced?c.) If you had 10 mol of C and SO₂ which reactant would be limiting?d.) What is the theoretical yield of CO, in moles if you react 10 moles of each reactant?

Part II: Limiting Reactant1. Consider the reaction: 5C+2SO2 → CS₂ + 4COIa.) If you-example-1
User Mittal Varsani
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Answer

a.) The moles of CO that would be produced = 8 moles

b.) The moles of CO that would be produced = 20 moles

c.) The limiting reactant is Carbon

d.) Theoretical yield of CO = 8 moles

Step-by-step explanation

Consider the given reaction: 5C + 2SO₂ → CS₂ + 4CO

a.) If you had 10 mol of Carbon, the moles of CO that would be produced is calculated using the mole ratio from the equation as follows


\begin{gathered} 5\text{ }mol\text{ }C=4\text{ }mol\text{ }CO \\ \\ 10\text{ }mol\text{ }C=x\text{ }mol\text{ }CO \\ \\ x=\frac{10\text{ }mol\text{ }C}{5\text{ }mol\text{ }C}*4\text{ }mol\text{ }CO \\ \\ x=8\text{ }mol\text{ }CO \end{gathered}

The moles of CO that would be produced = 8 moles

b.) If you had 10 mol of sulfur dioxide, the moles of CO that would be produced is calculated using the mole ratio from the equation as follows


\begin{gathered} 2\text{ }mol\text{ }SO_2=4\text{ }mol\text{ }CO \\ \\ 10mol\text{ }SO_2=x\text{ }mol\text{ }CO \\ \\ x=\frac{10\text{ }mol\text{ }SO_2}{2\text{ }mol\text{ }SO_2}*4\text{ }mol\text{ }CO \\ \\ x=20\text{ }mol\text{ }CO \end{gathered}

The moles of CO that would be produced = 20 moles

c.) From the equation of reaction; the mole ratio of C to SO₂ is 5:2.

This implies 5 moles of C requires 2 moles of SO₂

If you now had 10 mol of C and SO₂, it implies 10 mol of C will require just 4 mol of SO₂. Therefore, C is the limiting reactant because it will be the first reactant to be completely consumed.

d.) Since the limiting reactant is C (from part c); which determines when the reaction goes to completion and the amount of CO produced, then from part (a), 10 mol C produced 8 mol CO.

Therefore, the theoretical yield of CO = 8 moles

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