Answer to Question #135568 in General Chemistry for Misheck Mwilima Munyandi

Question

⦁ What mass of CO is required to react with 25.13gn of Fe2O3 according to the chemical reaction below:

  Fe2O3  +    CO           Fe   + CO2

⦁ How many molecules of NH3 are produced by the reaction of 7 Mols of Ca(OH)2 according to the following. `

(NH4)2SO4 + Ca(OH)2    NH3 + CaSO4  +  H2O

⦁

⦁ Calculate the energy required to excite the hydrogen electron from n= 1 to n = 2. Also calculate the wavelength of light that must be absorbed by hydrogen atom in its ground state to reach this excited state.  

Answer 1

I. The balanced reaction equation can be written as follows:

 Fe₂O₃  +    3CO      $\rightarrow$      2Fe   + 3CO₂.

According to the stoichiometry of the reaction, 1 mol of Fe₂O₃ reacts with 3 mol of CO:

$n(Fe_2O_3) = \frac{n(CO)}{3}$ .

The number of the moles of Fe₂O₃ in 25.13 g can be calculated using its molar mass $M$ =159.69 g/mol:

$n(Fe_2O_3) = \frac{m}{M} = \frac{\text{25.13 g}}{159.69 \text{ g/mol}} = 0.1574$ mol.

Then, the number of the moles of CO that take part in the reaction is:

$n(CO) = 3n(Fe_2O_3) = 3·0.1574 = 0.4721$ mol.

Finally, the mass of CO is (the molar mass of CO is 28.01 g/mol):

$m(CO) = M·n = 28.01\text{ g/mol}·0.4721\text{ mol}$

 $m(CO) = 13.22$ g.

II. Again, let's write the balanced reaction equation:

(NH₄)₂SO₄ + Ca(OH)₂ $\rightarrow$ 2NH₃ + CaSO₄  +  2H₂O.

As one can see, when 1 mol of Ca(OH)₂ reacts, 2 mol of NH₃ is produced:

$n(Ca(OH)_2) = \frac{n(NH_3)}{2}$ .

Therefore, the number of the moles of NH₃ produced is the double of the number of the moles of calcium hydroxide that react:

$n(NH_3) = 2n(Ca(OH)_2) = 14$ mol.

According to the definition of the mole, there are N_A = 6.022·10²³ constituent particles in 1 mol of the substance. Then, the number of the molecules of NH₃ in 14 mol is:

$N = n·N_A = 14·6.022·10^{23} = 8.43·10^{24}$ .

III. In order to calculate the energy of an electronic level $n$ in the atom of hydrogen, one may use the following formula:

$E = -\frac{E_0}{n^2}$ , where $E_0 = 13.6$ eV.

Therefore, the energy difference between the level 1 and 2 is:

$E_2 - E_1 = -E_0(\frac{1}{4} -1) = 13.6·3/4 = 10.2$ eV.

The energy level n=1 is the lowest and is the ground state of the hydrogen atom. Therefore, the wavelength of light needed to be absorbed by the atom in order to be excited is simply the eV converted to nm:

$\lambda = 121.6$ nm.

The wavelength can also be calculated using Rydberg formula:

$\frac{1}{\lambda} = R_H (\frac{1}{n_1^2}-\frac{1}{n_2^2})$ , where $R_H = 1.09677583·10^7$ m^-1 and which gives the same results.

Answer: I. 13.22 g, II. 8.43·10²⁴ molecules, III. 10.2 eV and 121.6 nm.