Respuesta :
Answer: the percent yield of this reaction was 84.7% and the best option to answer the question is the last one (letter E, 85.00%)
Explanation:
The question requires us to determine the percent yield of a reaction, given the amount of product obtained, the chemical equation and the amount of reactants used.
The following information was provided by the question:
mass of Al2O3 produced = 16.0 g
mass of Al used = 10.0 g
mass of O2 used = 19.0 g
balanced chemical equation:
[tex]4Al+3O_2\rightarrow2Al_2O_3[/tex]To solve this problem, we'll need to go through the following steps:
1) Calculate the number of moles used of each reactant;
2) determine the limiting reactant from the stoichiometry of the reaction and the amount of reactants used;
3) calculate the theoretical yield of the reaction, or, in other words, the amount of Al that should be produced, considering the limiting reactant;
4) calculate the percent yield of the reaction.
Next, we'll go through these steps to solve the problem:
1) Calculating the number of moles of each reactant
We can use the following equation to determine the amount of moles of Al and O2 that were used in the reaction:
[tex]n=\frac{m}{MM}[/tex]where n is the number of moles (in mol), m is the mass of the sample (in grams) and MM is the molar mass of the compound (in g/mol).
Knowing that the molar masses of Al and O2 are 26.98 and 31.98 g/mol, respectively, we can calculate the number of moles of each reactant as:
[tex]\begin{gathered} n_{Al}=\frac{10.0g}{26.98g/mol}=0.371mol\text{ Al} \\ n_{O_2}=\frac{19.0g}{31.98g/mol}=0.594mol\text{ }O_2 \end{gathered}[/tex]
Therefore, 0.371 and 0.594 moles of Al and O2 were used in the reaction, respectively.
2) Determining the limiting reactant.
From the balanced chemical equation, we can see that 4 moles of Al are necessary to react with 3 moles of O2. Thus, we can determine how many moles of O2 would be necessary to react with 0.371 moles of Al:
4 mol Al ------------------- 3 mol O2
0.371 mol Al ------------- x
Solving for x, we'll have:
[tex]x=\frac{(3mol\text{ }O_2)\times0.371mol\text{ Al\rparen}}{(4mol\text{ Al\rparen}}=0.278mol\text{ }O_2[/tex]Therefore, 0.278 moles of O2 would be necessary to react with the used amount of Al (0.371 mol). Since the actual amount of O2 used is greater than the necessary amount, we can say that O2 is the excess reactant and Al is the limiting reactant.
3) Calculating the theoretical amount of Al2O3 produced
Now that we know that Al was the limiting reactant in this reaction, we can determine how much Al2O3 should be produced in the reaction.
From the balanced chemical equation, we can see that 4 moles of Al are necessary to produce 2 moles of Al2O3. Thus, we can write:
4 mol Al --------------------- 2 mol Al2O3
0.371 mol Al --------------- y
Solving for y, we'll have:
[tex]y=\frac{(2mol\text{ A}l_2O_3)\times(0.371mol\text{ Al\rparen}}{(4mol\text{ Al\rparen}}=0.186mol\text{ A}l_2O_3[/tex]Therefore, with the amount of Al used, 0.186 moles of Al2O3 would be produced.
We can convert this amount in mass of Al2O3 using its molar mass (MM = 101.96 g/mol):
[tex]\begin{gathered} n=\frac{m}{MM}\rightarrow m=n\times MM \\ n_{Al_2O_3}=(0.186mol)\times(101.96g/mol)=18.91g \end{gathered}[/tex]Therefore, 18.9 g of Al2O3 should be obtained from the given mass of Al given.
4) Calculating the percent yield of the reaction
Note that the amount of Al2O3 expected, from the amount of reactants given, was 18.9g, but only 16.0g of the product was obtained. We can calculate the percent yield of a reaction using the following equation:
[tex]\begin{gathered} percent\text{ yield = }\frac{actual\text{ yield \lparen g\rparen}}{theoretical\text{ yield \lparen g\rparen}}\times100\% \\ \\ \%yield=\frac{16.0g}{18.9g}\times100\%=84.7\% \end{gathered}[/tex]Therefore, the percent yield of this reaction was 84.7% and the best option to answer the question is the last one (letter E, 85.00%).
