Answer:
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.
