How many grams of propane (C3H8) will react with 3.29 L of O2 at 1.05 atm and - 34° C? (Balance & use this equation: __ C3H8 + __ O2 → __CO2 + __ H2O)

1.55g of propane, C3H8

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

C3H8 + 5O2 - > 3CO2 + 4H2O

Step 2:

Data obtained from the question. This include the following:

Volume (V) of O2 = 3.29L

Pressure (P) = 1.05 atm

Temperature (T) = - 34°C = - 34°C + 273 = 239K

Number of mole O2 = ... ?

Gas constant (R) = 0.082atm. L/Kmol

Step 3:

Determination of the number of mole of O2 that reacted.

The number of mole of O2 that reacted can be obtained by using the ideal gas equation as follow:

PV = nRT

Divide both side by RT

n = PV / RT

n = (1.05 x 3.29) / (0.082 x239)

n = 0.176 mole

Therefore, 0.176 mole of O2 was used in the reaction.

Step 4:

Determination of the number of mole of C3H8 needed to react with 0.176 mole of O2.

This can be obtained as follow:

From the balanced equation above,

1 mole of C3H8 reacted with 5 moles of O2.

Therefore, Xmol of C3H8 will react with 0.176 mole of O2 i. e

Xmol of C3H8 = 0.176/5

Xmol of C3H8 = 0.0352 mole

Step 5:

Conversion of 0.0352 mole of C3H8 to grams.

This is illustrated below:

Mole of C3H8 = 0.0352 mole

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

Mass of C3H8 = ... ?

Mass = mole x molar Mass

Mass of C3H8 = 0.0352 x 44

Mass of C3H8 = 1.55g.

Therefore, 1.55g of propane, C3H8 were used in the reaction.