1.00 L of pure methanol, CH3OH, is reacted with 1.00 kg of oxygen according to t

Question

1.00 L of pure methanol, CH3OH, is reacted with 1.00 kg
of oxygen according to the following reaction:

2 CH3OH(l) + 3 O2(g) 2 CO2(g) + 4 H2O(l)

Determine the amount of carbon dioxide in grams that is
produced in the reaction, assuming 83.5% yield based on the
limiting reactant. The density of methanol is 0.793 g/ml.

First determine the limiting reagent. Then determine the
theoretical yield. Then calculate the actual yield.

1) Convert liters of methanol to grams of methanol using
the density. Then calculate moles of methanol by dividing
by the molecular weight of methanol.
2) Calculate moles of oxygen using the mass and molecular
weight of O2.
3) Using the balanced equation, calculate moles of CO2
formed from your moles of CH3OH, and moles of CO2 formed
from your moles of O2. Whichever one is less is your
limiting reagent.
4) Multiply moles of CO2 formed by the molecular weight of
CO2 to get grams of CO2. Then multiply by 0.835 to get the
actual yield in grams.

Explanation / Answer

2 CH3OH(l) + 3 O2(g) 2 CO2(g) + 4 H2O(l)
density = mass / Volume mass = density * Volume          = 0.793 g /mL* 1000 mL          = 793 g Molar mass of CH3OH = 12 + 4 * 1 + 16 = 32 g Molar mass od O2 = 2 * 16 = 32 g Molar mass of CO2 = 12 + 2 * 16 = 44 g 2 * 32 g of CH3OH reacts with 3 * 32 g of O2 793 g of CH3OH reacts with X g of O2 X = ( 3 * 32 * 793 ) / 64     = 1189.5 g So, 1189.5 - 1000 = 189.5 g of O2 left unreacted 2 * 32 g of CH3OH produces 2 * 44 g of CO2 793 g of CH3OH producesY g of CO2 Y = ( 2*44*793 ) / 64     = 1090.375 g of CO2 83.5 % of 1090.375 = (1090.375 * 83.5 ) / 100                                = 910.46 g ofCO2 mass = density * Volume          = 0.793 g /mL* 1000 mL          = 793 g Molar mass of CH3OH = 12 + 4 * 1 + 16 = 32 g Molar mass od O2 = 2 * 16 = 32 g Molar mass of CO2 = 12 + 2 * 16 = 44 g 2 * 32 g of CH3OH reacts with 3 * 32 g of O2 793 g of CH3OH reacts with X g of O2 X = ( 3 * 32 * 793 ) / 64     = 1189.5 g So, 1189.5 - 1000 = 189.5 g of O2 left unreacted 2 * 32 g of CH3OH produces 2 * 44 g of CO2 793 g of CH3OH producesY g of CO2 Y = ( 2*44*793 ) / 64     = 1090.375 g of CO2 83.5 % of 1090.375 = (1090.375 * 83.5 ) / 100                                = 910.46 g ofCO2 793 g of CH3OH producesY g of CO2 Y = ( 2*44*793 ) / 64     = 1090.375 g of CO2 83.5 % of 1090.375 = (1090.375 * 83.5 ) / 100                                = 910.46 g ofCO2

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.