Question

Solid aluminum and gaseous oxygen react in a combination reaction to produce aluminum oxide: 4Al (s) + 3O2 (g) → 2Al2O3 (s) In a particular experiment, the reaction of 2.5 g of Al with 2.5 g of O2 produced 3.5 g of Al2O3. The % yield of the reaction is ________. 3/18

a. 37
b. 66
c. 26
d. 74
e. 47

Answer 1

Answer: The percent yield of the reaction is 74 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$      .....(1)

$\text{Moles of aluminium}=\frac{2.5g}{27g/mol}=0.092mol$

For oxygen gas:

$\text{Moles of oxygen gas}=\frac{2.5g}{32g/mol}=0.078mol$

The chemical equation for the reaction of titanium and chlorine gas follows:

$4Al(s)+3O_2(g)\rightarrow 2Al_2O_3(s)$

By Stoichiometry of the reaction:

4 moles of aluminium reacts with 3 moles of oxygen.

So, 0.092 moles of aluminium reacts with = $\frac{3}{4}\times 0.092=0.069mol$ of oxygen

As, given amount of oxygen is more than the required amount. So, it is considered as an excess reagent.

Thus, aluminium is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

4 moles of aluminium produce = 2 moles of $Al_2O_3$

So, 0.092 moles of aluminium will produce = $\frac{2}{4}\times 0.092=0.046moles$ of $Al_2O_3$

Now, calculating the mass of aluminium oxide:

$\text{Mass of aluminium oxide}=moles\times {\text {molar mas}}=0.046mol\times 102g/mol=4.7g$

To calculate the percentage yield of titanium (IV) chloride, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield  = 3.5 g

Theoretical yield = 4.7 g

Putting values in above equation, we get:

$\%\text{ yield of reaction}=\frac{3.5g}{4.7g}\times 100\\\\\% \text{yield of reaction}=74\%$

Hence, the percent yield of the reaction is 74 %

Answer 2

The percent yield of the aluminum oxide reaction has been 74%.

From the balanced chemical equation of aluminum with oxygen, 4 moles of Al reacts with 3 moles of oxygen, to give 2 moles of aluminum oxide.

The moles of reactants and products in the reaction can be given by:

Moles = $\rm \dfrac{weight}{molecular\;weight}$

Moles of 2.5g Al = $\rm \dfrac{2.5}{27}$

Moles of Al = 0.092 mol

Moles of oxygen = $\rm \dfrac{2.5}{32}$

Moles of oxygen = 0.078 mol

From the balanced equation:

4 moles Al = 3 moles oxygen

0.092 moles Al = 0.069 moles of oxygen .

Since the available oxygen has been in access, Aluminum has been the limiting reagent,

Thus, the moles of Aluminum oxide produced has been:

4 moles Al = 2 moles Aluminum oxide

0.092 moles Al = 0.046 mol of Aluminum oxide

The mass of 0.046 mol aluminum oxide:

Mass = Moles × Molecular weight

Mass of 0.046 mol Aluminum oxide = 0.046 mol × 102 g/mol

Mass of 0.046 mol Aluminum oxide = 4.7 grams.

The theoretical yield of Aluminum oxide = 4.7 grams.

The actual yield of Aluminum oxide = 3.5 grams.

The %yield can be given as:

% yield = $\rm \dfrac{Actual\;yield}{Theoretical\;yield}\;\times\;100$

% yield of the reaction = $\rm \dfrac{3.5}{4.7}\;\times\;100$

% yield of reaction = 74%

The percent yield of the aluminum oxide reaction has been 74%.

For more information about the percent yield, refer to the link:

https://brainly.com/question/25781371