Using 18.8L, calculate the volume (in L) of water vapour that should be produced by the reaction of 35.0 g

A. 18.8L

B. 75.2L of H2O.

Explanation:

A. Determination of the volume of 35g of C3H8.

Date obtained from the question include the following:

Mass of C3H8 = 35g

Temperature (T) = 40°C

Pressure (P) = 110KPa

Volume (V) =..?

Next, we shall determine the number of mole (n) in 35g of C3H8. This is illustrated below:

Molar mass of C3H8 = (3x12) + (8x1) = 44g/mol

Mass of C3H8 = 35g

Mole of C3H8 =..?

Mole = mass /molar mass

Mole of C3H8 = 35/44

Mole of C3H8 = 0.795 mole

Finally, we shall determine the volume of 35g of C3H8 as follow:

Temperature (T) = 40°C = 40°C + 273 = 313K

Pressure (P) = 110KPa

Number of mole (n) = 0.795 mole

Gas constant (R) = 8.314 KPa.L/Kmol

Volume (V) =..?

PV = nRT

110 x V = 0.795 x 8.314 x 313

Divide both side by 110

V = (0.795 x 8.314 x 313)/110

V = 18.8L

Therefore, the volume of 35g of C3H8 under the conditions given is 18.8L

B. Determination of the volume of water vapour produced by the reaction of 35g of propane, C3H8.

From the calculations made in (A) above, 35g of C3H8 is equivalent to 18.8L of C3H8.

Thus, we can obtain the volume of water vapour produced as follow:

C3H8 + 5O2 —> 3CO2 + 4H2O

From the balanced equation above,

1L of C3H8 reacted to produce 4L of H2O.

Therefore, 18.8L of C3H8 will react to produce = (18.8 x 4)/1 = 75.2L of H2O.

Therefore, 75.2L of H2O were produced from the reaction.