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what volume of water at 742 torr and 35°c would be produced when 45 g of oxygen reacts with excess ethane

User NirKa
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1 Answer

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17 votes

The ethane will react with oxygen to produce carbon dioxide and water:

2 C₂H₆ + 7 O₂ ----> 4 CO₂ + 6 H₂O

We have to find the volume of water produced (we will consider that it is vaporized) from 45 g of oxygen with excess ethane. The first step is to convert those grams into moles of oxygen using the molar mass of it.

atomic mass of O = 16.00 amu

molar mass of O₂ = 2 * 16.00 g/mol

molar mass of O₂ = 32 g/mol

moles of O₂ = mass of O₂/molar mass of O₂

moles of O₂ = 45 g/(32 g/mol)

moles of O₂ = 1.41 moles

2 C₂H₆ + 7 O₂ ----> 4 CO₂ + 6 H₂O

According to the balanced equation, when 7 moles of O₂ (reacting with excess ethane) produces 6 moles of H₂O. So we can use that relationship to find the number of moles of H₂O produced by 1.41 moles of O₂.

moles of H₂O = 1.41 moles of O₂ * 6 moles of H₂O/(7 moles of O₂)

moles of H₂O = 1.21 moles

Now we can use the ideal gas law to find the volume of water.

P * V = n * R * T

R = 0.082 atm*L/(mol*K)

P = 742 torr

1 atm = 760 torr

P = 742 torr * 1 atm/(760 torr)

P = 0.976 atm

V = ?

n = 1.21 moles

T = 35 °C = (35 + 273.15) K

T = 308.15 K

We will find the answer to our problem by replacing those values in the ideal gas law and solving it for the volume:

P * V = n * R * T

V = n * R * T /P

V = 1.21 mol * 0.082 atm*L/(mol*K) * 308.15 K/(0.976 atm)

V = 31.3 L

Answer: 31.3 L of water is produced.

User Nclark
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