Question

In the chemical equation CH4+2O2→CO2+2H2O what is the total amount of mass that is represented by the reactants ,methane (CH4) and the oxygen (O2) in amu

Answer 1

Answer:

The reaction is CH

4 +2O 2

→CO2+(2H) (2O).

At room temperature, water will condense to form liquid water.

1 mole of methane reacts with 2 moles of oxygen to form 1 mole of oxygen.

10 ml (0.445 moles) of methane will react with 20 ml (0.89 moles) of oxygen to form 10 ml (0.445 moles) of carbon dioxide.

Hence, the resulting mixture will contain 10 ml of unreacted methane and 10 ml of carbon dioxide.

The total volume of the mixture will be 20 ml.

Explanation:

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Answer 2

Given,

Chemical reaction,

$CH_{4} + 2O_{2} --> CO_{2} +2H_{2}O$

To find,

The total mass of the reactants in the atomic mass unit (a.m.u)

Solution,

The atomic mass unit can be defined as 1/12th the mass of one atom of carbon-12.

On further solving, we get 1 a.m.u = $1.66*10^{-21}gm$

$CH_{4} + 2O_{2} --> CO_{2} +2H_{2}O$

We observe that the given equation is already balanced.

Therefore, we can take the coefficients for comparison.

1 mole of methane will react with two moles of atmospheric oxygen to produce two moles of water molecules and one mole of carbon dioxide.

1 molecule of methane will react with two molecules of atmospheric oxygen to produce two molecules of water molecules and one molecule of carbon dioxide.

Mass of one molecule of methane = $(12*1 +1*4 ) a.m.u$

                                                          $= 16 a.m.u$

Mass of one molecule of oxygen = $2*(16*2) a.m.u$

                                                          = $64 a.m.u$

Total mass of reactants = mass of methane + 2*mass of oxygen molecule

                                       $= 16 a.m.u + 64 a.m.u\\= 80 a.m.u$

Therefore, the weight of reactants is $80 a.m.u$