How many moles of O2 should be supplied to burn 1 mol of C3H8 (propane) molecules in a camping stove

1 answer:

4 0

The combustion of any hydrocarbon yields water and carbon dioxide. We will now construct a balanced equation:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Each mole of propane requires 5 moles of oxygen.

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The gaseous product of a reaction is collected in a 25.0-l container at 27

nataly862011 [7]

To find the number of moles of gas we can use the ideal gas law equation, we dont need to use the mass of gas given as we only have to find the number of moles
PV = nRT
P - pressure - 300.0 kPa
V - volume - 25.0 x 10⁻³ m³
n - number of moles
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in Kelvin - 27 °C + 273 = 300 K
substituting these values in the equation
300.0 kPa x 25.0 x 10⁻³ m³ = n x 8.314 Jmol⁻¹K⁻¹ x 300 K
n = 3.01 mol
number of mols of gas - 3.01 mol

4 0

1 year ago

in which sample of water do the molecules have the highest average kinetic energy 1)20. mL at 100.°C 2)40. mL at 80. °C 3)60. mL

miv72 [106K]

100°C because all the molecules are moving the fastest past each other

6 0

2 years ago

A gas that has a volume of 28 liters, a temperature of 45C, And an unknown pressure has its volume increased to 34 liters and it

patriot [66]

Answer:

P1 = 2.5ATM

Explanation:

V1 = 28L

T1 = 45°C = (45 + 273.15)K = 318.15K

V2 = 34L

T2 = 35°C = (35 + 273.15)K = 308.15K

P1 = ?

P2 = 2ATM

applying combined gas equation,

P1V1 / T1 = P2V2 / T2

P1*V1*T2 = P2*V2*T1

Solving for P1

P1 = P2*V2*T1 / V1*T2

P1 = (2.0 * 34 * 318.15) / (28 * 308.15)

P1 = 21634.2 / 8628.2

P1 = 2.5ATM

The initial pressure was 2.5ATM

3 0

1 year ago

Write a balanced half-reaction for the oxidation of liquid water H2O to aqueous hydrogen peroxide H2O2 in basic aqueous solution

nignag [31]

Answer : The balanced half-reaction in a basic solution will be,

$2OH^-(aq)\rightarrow H_2O_2(aq)+2e^-$

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Rules for the balanced chemical equation in basic solution are :

First we have to write into the two half-reactions.

Now balance the main atoms in the reaction.

Now balance the hydrogen and oxygen atoms on both the sides of the reaction.

If the oxygen atoms are not balanced on both the sides then adding water molecules at that side where the more number of oxygen are present.

If the hydrogen atoms are not balanced on both the sides then adding hydroxide ion $(OH^-)$ at that side where the less number of hydrogen are present.