Answer:
See explanation
Explanation:
The equation of the reaction is;
C 2 H 6 O(l) + 3 O 2 (g) → 2 CO 2 ( g ) + 3 H 2 O(l )
Next we have to determine the limiting reactant, this reactant gives the least number of moles of product.
Number of moles of C 2 H 6 O = mass/molar mass = 1.86g/ 46.07 g/mol = 0.04 mols
From the equation;
1 mol of ethanol yields 2 mols of CO2
0.04 moles of ethanol yields 0.04 * 2/1 = 0.08 mols of CO2
For water;
1 mol of ethanol yields 3 mols of water
0.04 moles of ethanol yields 3 * 0.04/1 = 0.12 mols of water
Also;
Number of moles of oxygen= 10g/32g/mol = 0.31 moles
3mols of O2 yields 2 moles of CO2
0.31 moles of O2 yields 0.31 * 2/3 = 0.21 moles of CO2
For water;
3 moles of O2 yields 3 moles of water
0.31 moles of O2 yields 0.31 * 3/3 = 0.31 moles of water
Hence ethanol is the limiting reactant.
From PV=nRT
Volume of CO2 is;
V = nRT/P
V = 0.08 * 0.082 *298/1 = 1.95 L
Volume of water;
V = nRT/P
V= 0.12 * 0.082 * 298/1
V= 2.93 L
Total volume of gases after reaction = 1.95 L + 2.93 L = 4.88 L
