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

1) $Fe_2O_3 +3CO \rightarrow Fe + 3CO2$

$25.13 g (Fe_2O_3)\times (\frac{1 mole (Fe_2O_3)}{159.69g(Fe_2O_3)}) \times( \frac{3 mol (CO)}{1 mol (Fe_2O_3)})\times (\frac{28.01 g (CO)}{1 mol (CO)}) = 13.22 g (CO)$

2) $(NH_4)_2SO_4+Ca(OH)_2\rightarrow 2NH_3 + CaSO_4+2H_2O$

$7 mol (Ca(OH)_2)\times (\frac{2 mol (NH_3)}{1 mol Ca(OH)_2}) \times (\frac{6.022\times 10^{23} molecules (NH_3)}{1 mole (NH_3)})=8\times 10^{24} molecules NH_3$

3) $\Delta E = -R_H (\frac{1}{n^2_{final}}-\frac{1}{n^2_{initial}})$

$\Delta E = -2.18\times 10^{-18} J\times (\frac{1}{2^2}-\frac{1}{1^2})=1.635\times 10^{-18} J$

$\Delta E = \frac{hc}{λ}$

$λ=\frac{hc}{\Delta E}=\frac{6.626\times10^{-34}J\times s \times 2.998\times 10^8 m/s}{1.635\times 10^{-18}J}=1.215\times10^{-7}m$