Question

Hi sir I got answers of questions 10) 48.88 gm of Alcl3On base of questions 10 please solve questions 11) please

Answer 1

The actual yield of aluminum chloride is 25.49 gm

The theoretical yield of aluminum chloride is 48.98 gm

Explanation

Given information

The mass of aluminum = 34.0 gm

The mass of chlorine = 39.0 gm

The next step is to write down the balanced chemical reaction

`2Al_((s))+3Cl_(2(g))\text{ }\rightarrow2AlCl_3`

The next step is to find the moles of aluminum and chlorine using the below formula

`\text{mole = }\frac{\text{ mass}}{\text{ molar mass}}`

According to the periodic table, the molar mass of aluminum is 26.981 g/mol and the molar mass of chlorine is 35.5 g/mol

`\begin{gathered} \text{Mole of Al = }(34)/(26.981) \\ \text{Mole of Al = }1.2601\text{ moles} \\ \\ \text{Mole of Cl = }(39)/(70.8) \\ \text{Mole of Cl = 0}.551\text{ moles} \end{gathered}`

The next step is to determine the limiting reagent from the number of moles by dividing the moles by the coefficient

From the reaction, you will see that aluminum has 2 moles and chlorine has 3 moles

`\begin{gathered} Al\text{ = }(1.2601)/(2) \\ Al\text{ = 0.6301 moles} \\ \\ Cl_2\text{ = }(0.551)/(3) \\ Cl_2\text{ = 0.1837 moles} \end{gathered}`

From the calculations above, you will see that chlorine has the least number of moles, and it is the limiting reagent

Since the limiting reagent is chlorine, then, we can now find the number of moles of aluminum chloride using a stoichiometry ratio

From the balanced reaction, 3 moles of chlorine yield 2 moles of aluminum chloride.

Let x be the number of moles of aluminum chloride

`\begin{gathered} 3\text{ moles }\rightarrow\text{ 2 moles } \\ 0.551\text{ }\rightarrow\text{ x moles} \\ \text{cross multiply} \\ 2\text{ }*\text{ 0.551 = 3 }*\text{ x} \\ 1.102\text{ = 3x} \\ \text{divide both sides by 3} \\ (1.102)/(3)\text{ = }\frac{\cancel{3}x}{\cancel{3}} \\ x\text{ = 0.3673 moles} \end{gathered}`

Therefore, the number of moles of aluminum chloride is 0.3673 moles

The next step is to find the mass of aluminum chloride using the below formula

`\text{mole = }\frac{\text{ mass}}{\text{ molar mass}}`

According to the periodic table, the molar mass of aluminum chloride is 133.34 g/mol

`\begin{gathered} 0.3673\text{ = }(mass)/(133.34) \\ cross\text{ multiply} \\ \text{mass = 0.3673 }*133.34 \\ \text{mass = 48.98 gm} \end{gathered}`

Therefore, the theoretical yield of aluminum chloride is 48.98 gm

The next step is to find the actual yield of the aluminum chloride by using the below formula

`\text{percentage yield = }\frac{actual\text{ yield }}{\text{theoretical yield}}`

Recall that, the percentage yield is 51%

`\begin{gathered} (51)/(100)\text{ = }\frac{\text{ actual yield}}{48.98} \\ \text{Cross multiply} \\ 51\text{ }*\text{ 48.98 = actual yield }*\text{ 100} \\ \text{Actual yield = }\frac{51\text{ }*\text{ 48.98}}{100} \\ \text{Actual yield = }(2548.98)/(100) \\ \text{Actual yield = 25.4898} \end{gathered}`

Therefore, the actual yield of the aluminum chloride is 25.49 gm