Question

The minimum amount of O (g) consumed per gram of the reactant is for the reaction:
(Given atomic mass: Fe = 56, O = 16, Mg = 24, P = 31, C =12, H = 1).

A. $\text{C}_{3}^{}\text{H}_{8}^{}(\text{g})+5\text{O}_{2}^{}(\text{g})\to 3\text{CO}_{2}^{}(\text{g})+4\text{H}_{2}^{}\text{O}(\text{l})$
B. ${{P}_{4}}(s)+5O{}_{2}(g)\to {{P}_{4}}{{O}_{10}}(s)$
C. $4Fe(s)+3{{O}_{2}}(g)\to 2Fe{{O}_{3}}(s)$
D. $2Mg(s)+{{O}_{2}}(g)\to 2MgO(s)$

Answer 1

. We have to calculate how many grams of oxygen required oxidizing the one gram of each chemical with the following formula.
One gram of chemical = $=\dfrac{\text{number of moles of the oxygen }\\!\\!\times\\!\\!\text{ molecular weight of the oxygen}}{\text{number of moles of the chemical }\\!\\!\times\\!\\!\text{ molecular weight of the chemical}}$

Complete step by step answer:
- In the question it is asked to find the minimum amount of oxygen consumed per gram of the reactant for the reactions in the given options.
- Coming to the given options, option A, $\text{C}_{3}^{}\text{H}_{8}^{}(\text{g})+5\text{O}_{2}^{}(\text{g})\to 3\text{CO}_{2}^{}(\text{g})+4\text{H}_{2}^{}\text{O}(\text{l})$ .
- The molecular weight of the reactant in option A is 44.
- Number of moles of the reactant is one.
- Molecular weight of the oxygen is 32.
- Number of moles of the oxygen in option A is five.
- By using the below formula we can calculate the minimum amount of oxygen required by the reactant chemical in the option A, $\text{C}_{3}^{}\text{H}_{8}^{}(\text{g})+5\text{O}_{2}^{}(\text{g})\to 3\text{CO}_{2}^{}(\text{g})+4\text{H}_{2}^{}\text{O}(\text{l})$ .
One gram of chemical (propane)

& =\dfrac{\text{number of moles of the oxygen }\\!\\!\times\\!\\!\text{ molecular weight of the oxygen}}{\text{number of moles of the chemical }\\!\\!\times\\!\\!\text{ molecular weight of the chemical}} \\\ & =\dfrac{5\times 32}{1\times 44} \\\ & =3.63g \\\ \end{aligned}$$ \- Therefore to oxidize one gram of propane there is a need of 3.63 g of oxygen. \- Coming to the given options, option B, ${{P}_{4}}(s)+5O{}_{2}(g)\to {{P}_{4}}{{O}_{10}}(s)$ \- The molecular weight of the reactant in option B is 124. \- Number of moles of the reactant is one. \- Molecular weight of the oxygen is 32. \- Number of moles of the oxygen in option B is five. \- By using the below formula we can calculate the minimum amount of oxygen required by the reactant chemical in option B. One gram of chemical (phosphorous) $$\begin{aligned} & =\dfrac{\text{number of moles of the oxygen }\\!\\!\times\\!\\!\text{ molecular weight of the oxygen}}{\text{number of moles of the chemical }\\!\\!\times\\!\\!\text{ molecular weight of the chemical}} \\\ & =\dfrac{5\times 32}{1\times 124} \\\ & =1.29g \\\ \end{aligned}$$ \- Coming to the given options, option C, $4Fe(s)+3{{O}_{2}}(g)\to 2Fe{{O}_{3}}(s)$ \- The molecular weight of the reactant in option C is 56. \- Number of moles of the reactant is 4. \- Molecular weight of the oxygen is 32. \- Number of moles of the oxygen in option C is three. \- By using the below formula we can calculate the minimum amount of oxygen required by the reactant chemical in the option C. One gram of chemical (Iron) $$\begin{aligned} & =\dfrac{\text{number of moles of the oxygen }\\!\\!\times\\!\\!\text{ molecular weight of the oxygen}}{\text{number of moles of the chemical }\\!\\!\times\\!\\!\text{ molecular weight of the chemical}} \\\ & =\dfrac{3\times 32}{4\times 56} \\\ & =0.42g \\\ \end{aligned}$$ \- Coming to the given options, option D, $2Mg(s)+{{O}_{2}}(g)\to 2MgO(s)$ \- The molecular weight of the reactant in option D is 24. \- Number of moles of the reactant is 2. \- Molecular weight of the oxygen is 32. \- Number of moles of the oxygen in option D is one. \- By using the below formula we can calculate the minimum amount of oxygen required by the reactant chemical in the option C. One gram of chemical (Magnesium) $$\begin{aligned} & =\dfrac{\text{number of moles of the oxygen }\\!\\!\times\\!\\!\text{ molecular weight of the oxygen}}{\text{number of moles of the chemical }\\!\\!\times\\!\\!\text{ molecular weight of the chemical}} \\\ & =\dfrac{1\times 32}{2\times 24} \\\ & =0.66g \\\ \end{aligned}$$ \- Therefore the least amount of oxygen is consumed by Iron (0.42 g). **So, the correct answer is “Option C”.** **Note:** The reaction of any chemical with oxygen is called oxidation. During oxidation almost all metals lose their electrons and will get oxidized and at the same time oxygen will accept the electrons and get reduced itself.