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3 years ago

2.3g of sodium metal is burned completely in oxygen of air.

Chemistry

1 answer:

Nina [5.8K]3 years ago

4 0

Answer:

mass of Na2O= 3.1 grams

volume of O2= 0.56 L

Explanation:

First what you need to do is establish the balanced equation for the reaction that they're giving you. In this case would be:

4Na + O2 > 2Na2O

And then, you can read the stequiometry of the reaction:

4 moles of Na react with one mole of oxygen to produce 2 moles of sodium oxide.

So now you know that you have 2.3 grams of sodium so you have to obtain the quantity of moles that are present in 2.3 grams of sodium. The molar mass the (the grams by which you will have exactly one mole) so all you have to do is divide 2.3g ÷ 23g/mol= 0.1 mol

So here you have 0.1 mol of sodium, and you know for a fact that for each 4 moles of sodium, one mole of oxygen reacts, so you simply divide the quantity of moles of sodium you have by the 4 moles of oxygen present in the balanced equation.

0.1 moles ÷ 4 moles = 0.025.

So for each 0.1 moles of sodium that react, 0.025 moles of Oxygen react.

Now what you need to do is to find the limitant reactor (the reactor that's according to the comparison to the balanced equation is in a smaller ratio to the other element, that would be in excess.

In this case, both elements are in a ratio that respects the balanced equation so neither of them is in excess or is limitant.

Now, to determine the amount of Na2O, you simply know that one mole of oxygen react to produce to moles of sodium oxide, So all you have to do is establish the factor of conversion:

0.025 moles O2 x 2 moles Na2O/ 1 mol O2= 0.05 moles of Na2O

Then you convert that to grams by multiplying 0.05 by the molar mass of Sodium oxide and it will be:

0.05 mol Na2O x 62 g/mol= 3.1 grams of Na2O

Now what you need to do is find the volume by which O2 is reacting in the reaction so you can use the equation of the ideal gas law.

PV=nRT at standard temperature and pressure (273 K and 1 Atm)

P=Pressure

V=Volume

n=Number of atoms

R=constant of gases (0.082 Atm L/ mol K)

T=Temperature

V=nRT/P (because you need to find the volume)

V= (0.025 mol)(0.082)(273 K)/1 atm

V=0.56 L

And there you go :)

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a) 1.12 MeV / nucleon

b) 5.62 MeV / nucleon

c) 8.80 MeV / nucleon

d) 8.56 MeV / nucleon

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Explanation:

Binding energy = ( Δm * 931.5 ) MeV

Binding energy per nucleon = Binding energy in / Number of nucleon

a) ²H = 1 neutron , 1 proton = 2 nucleons

Given that the theoretical mass = 2.0141 u

Actual mass = 1.0078 u + 1.0087 u = 2.0165 u

Δm = 2.0165 u - 2.0141 u = 2.4 * 10^-3 u

∴ Binding energy per nucleon = ( 2.4 * 10^-3 * 931.5 ) MeV / 2 nucleons

= 1.12 MeV / nucleon

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Actual mass = ( 3 * 1.0078 ) + ( 4 * 1.0087 ) = 7.0582 u

Δm = ( 7.0582 u - 7.0160 u ) = 0.0422 u

∴ Binding energy per nucleon = ( 0.0422 * 931.5 ) / 7

= 5.62 MeV / nucleon

C) ⁶²Ni = 28 protons , 34 neutrons = 62 nucleons

Theoretical mass = 61.9283 u

Actual mass = ( 28 * 1.0078 ) u + ( 34 * 1.0087 ) u

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Δm = 0.5859 u

∴ Binding energy per nucleon = ( 0.5859 * 931.5 ) / 62

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D) ¹¹⁰Cd = 48 protons , 62 neutrons = 110 nucleons

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∴ Binding energy per nucleon = ( 1.0108 * 931.5 ) / 110

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