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.

b

explanation:

mitosis

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.