Answer:
9.94 mL, the volume of ethanol needed
Explanation:
The reaction is:
C₂H₅OH(l) + 3 O₂(g) → 2 CO₂(g) + 3 H₂O(l)
We convert the mass of the formed product to moles:
15 g . 1mol / 44g = 0.341 moles
2 moles of dioxide are produced by 1 mol of ethanol, in order to stoichiometry.
Therefore, 0.341 moles of CO₂ must be produced by (0.341. 1) / 2 = 0.1705 moles of alcohol.
We convert the moles to mass, and then, the mass to volume by the use of density.
0.1705 mol . 46 g / 1 mol = 7.84 g of ethanol
Ethanol density = Ethanol mass /Ethanol volume
Ethanol volume = Ethanol mass /Ethanol density → 7.84 g /0.789 g/mL =
9.94 mL
Answer:
We need 9.95 mL of ethanol
Explanation:
Step 1: Data given
Density ethanol = 0.789 g/mL
Mass CO2 produced = 15.0 grams
Step 2: The balanced equation
C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l)
Step 3: Calculate moles CO2
Moles CO2 = mass CO2 / molar mass CO2
Moles CO2 = 15.0 grams / 44.01 g/mol
Moles CO2 = 0.341 moles
Step 4: Calculate moles ethanol
For 1 mol ethanol, 3 moles we need O2 to produce 2 moles CO2 and 3 moles H2O
For 0.341 moles CO2 we need 0.341 /2 = 0.1705 moles ethanol
Step 5: Calculate mass ethanol
Mass ethanol = moles ethanol * molar mass ethanol
Mass ethanol = 0.1705 moles * 46.07 g/mol
Mass ethanol = 7.85 grams
Step 6: Calculate volume ethanol
Volume ethanol = mass / density
Volume = 7.85 grams / 0.789 g/mL
Volume= 9.95 mL
We need 9.95 mL of ethanol
