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Chapter 18: Problem 49

Sodium stearate $\left(\mathrm{C}_{18} \mathrm{H}_{35} \mathrm{O}_{2} \mathrm{Na}\right)$ is the most common soap. Assume that the stearate anion undergoes aerobic decomposition in the following manner: $$ \begin{aligned} \mathrm{C}_{18} \mathrm{H}_{35} \mathrm{O}_{2}^{-}(a q)+26 \mathrm{O}_{2}(a q) & \longrightarrow \\ & 17 \mathrm{CO}_{2}(a q)+17 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{HCO}_{3}^{-}(a q) \end{aligned} $$ What is the total mass of \(\mathrm{O}_{2}\) required to biodegrade $3.0 \mathrm{~g}$ of this substance?

Short Answer

Expert verified
The total mass of O₂ required to biodegrade 3.0 g of sodium stearate is \(8.64 \: \text{g}\).

Step by step solution

01

Find the moles of sodium stearate

First, we need to find the molar mass of sodium stearate (C₁₈H₃₅O₂Na). To do this, we add the molar masses of each element in the compound: Molar mass of C₁₈H₃₅O₂Na = 18×(12.01 amu) + 35×(1.01 amu) + 2×(16.00 amu) + 1×(22.99 amu) = 287.46 g/mol Now, we can calculate the moles of sodium stearate: moles of C₁₈H₃₅O₂Na = (mass of sodium stearate)/(molar mass of sodium stearate) moles of C₁₈H₃₅O₂Na = (3.0 g)/(287.46 g/mol) = 0.0104 mol
02

Balance the given reaction

The given reaction is already balanced, so we can use the stoichiometric coefficients to find the moles of O₂ required to react with sodium stearate: C₁₈H₃₅O₂⁻ + 26 O₂ → 17 CO₂ + 17 H₂O + HCO₃⁻ According to the balanced equation, 26 moles of O₂ are needed for every 1 mole of sodium stearate.
03

Calculate the moles of O₂ required

Using the stoichiometric coefficients from Step 2, we can calculate the moles of O₂ needed: moles of O₂ = (moles of C₁₈H₃₅O₂Na) × (26 moles O₂ / 1 mole C₁₈H₃₅O₂Na) moles of O₂ = 0.0104 mol C₁₈H₃₅O₂Na × (26 moles O₂/1 mole C₁₈H₃₅O₂Na) = 0.270 mol O₂
04

Calculate the mass of O₂ required

Now, we can find the mass of O₂ required by multiplying the moles of O₂ by the molar mass of O₂: mass of O₂ = moles of O₂ × molar mass of O₂ mass of O₂ = 0.270 mol × 32.00 g/mol = 8.64 g So, 8.64 g of O₂ is required to biodegrade 3.0 g of sodium stearate.

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