1.86 g of ethanol reacts with 10.0 g of oxygen. What is the total volume of gas present (in L) after the reaction is complete, assuming the reaction takes place at 1.00 atm and 25oC?

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

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