#1) calculate the volume of water vapour that is produced from the combustion of 15.0 g of ethylene at 25°c and 100 kpa.

V = 24.7 L

Explanation:

Given data:

Mass of ethylene = 15.0 g

Temperature = 25°C (25+273 = 298 K)

Pressure = 100 KPa ( 100/101 = 0.99 atm)

Volume of water produced = ?

Solution:

Chemical equation:

C₂H₄ + 3O₂  →  2CO₂ + 2H₂O

Number of moles of ethylene:

Number of moles = mass / molar mass

Number of moles = 15.0 g/ 28.05 g/mol

Number of moles = 0.5 mol

Now we will compare the moles of ethylene with water.

             C₂H₄           :               H₂O

               1                :                  2

               0.5           :               2/1×0.5 = 1 mol

Volume of water:

PV = nRT

V = nRT/P

V = 1 mol × 0.0821 atm. L/mol.K ×298 K / 0.99 atm

V = 24.5 atm.L /0.99 atm

V = 24.7 L

the equation for synthesis of water is as follows;

here two moles of hydrogen are used with one mole of oxygen which gives one molecule of water as a product. but shana misinterpreted 2 moles as 2 grams in this equation.