10. Consider the following reaction between isopropyl alcohol and oxygen C3H50 (

Question

10. Consider the following reaction between isopropyl alcohol and oxygen C3H50 (g) O2 (g) CO2 (g) H20 (a) What type of reaction is this? (b) Balance the equation for this reaction. You can use the space below to write the equation or fill it in above, but make sure your final answer is clearly labeled. (c) If you start with 25.0 mL ofC3Hso, and the density of isopropyl alcohol is 0.786 glmL, what mass of O2 will be required to completely react with the isopropyl alcohol? (d) If 25.0 mL of C3Hso react with 75.0 g of O2, which substance will be the limiting reactant?

Explanation / Answer

(a) It is combustion reaction as the organic compound is being burned in air (O2)

(b)

2 C3H8O (g) + 9 O2 (g) -------------> 6 CO2 (g) + 8 H2O (g)

(c)

Mass of isopropyl alcohol = density * volume = 0.786 * 25.0 = 19.65 g.

Moles of isopropyl alcohol = mass / molar mass = 19.65 / 60 = 0.328 mol

From the balanced equation,

2 mol of isopropyl alcohol requires 9 mol of O2

then, 0.328 mol of isopropyl alcohol requires 0.328 * 9 / 2 = 1.476 mol of O2

Mass of O2 required = 1.476 * 32 = 47.2 g.

(d)

From (c) 25.0 mL of isopropyl alcohol requires 47.2 g. of O2. If we 75.0 g. of O2 it is excess reagent and isopropyl alcohol is limiting reagent.

(e)

From the balanced equation,

2 mol isopropyl alcohol = 6 mol CO2

then,

0.328 mol isopropyl alcohol = 6 * 0.328 / 2 = 0.984 mol of CO2

Theretical yield of CO2 = 0.984 * 44 = 43.3 g.

(f)

% yield = (actual yield / theoretical yield) * 100 = (40/43.3) * 100 = 92.4 %

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