Question

How many grams of AlCl3 can be produced from the reaction that occurs between 94g Al and 197g of Cl2

Answer 1

The limiting reactant is chlorine

The mass of AlCl3 is 246.41 grams

Explanation

Given information

The mass of aluminum = 94 grams

The mass of chlorine = 197 grams

Firstly, we need to write a balanced equation of the reaction

`2Al\text{ + 3Cl}_2\text{ }\rightarrow\text{ 2AlCl}_3`

From the reaction above, you will see that 2 moles of aluminum react with 3 moles of aluminum

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

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

Recall, that the molar mass of aluminum and chlorine is 27 g/mol and 71 g/mol

For aluminum, the number of moles is calculated below as

`\begin{gathered} mole\text{ = }(94)/(27) \\ mole\text{ = 3.48 moles} \end{gathered}`

For chloride, the number of moles can be calculated below as

`\begin{gathered} mole\text{ = }(197)/(71) \\ mole\text{ = 2.775 moles} \end{gathered}`

Since we have gotten the number of moles of each reactant, therefore, we can now determine the limiting reactant by dividing the number of moles of each reactant by its coefficient.

The coefficient of aluminum in the reaction is 2 and the coefficient of chloride in the reaction is 3

`\begin{gathered} For\text{ Al} \\ (3.48)/(2)\text{ = 1.74 moles} \\ \\ For\text{ Cl} \\ (2.775)/(3)\text{ = 0.925 moles} \end{gathered}`

From the above calculations, you will see that chlorine has the least number of moles which is 0.925 moles.

Therefore, the limiting reactant is chlorine.

The next step is to find the number of moles of aluminum chloride using a stoichiometry ratio.

`\begin{gathered} \text{ 3 moles of chlorine }\rightarrow\text{ 2 moles of aluminum chloride} \\ 2.775\text{ moles }\rightarrow\text{ x moles of aluminum chloride} \\ cross\text{ multiply} \\ 3*\text{ x = 2 }*\text{ 2.775} \\ 3x\text{ = 5.544} \\ Divide\text{ both sides by 3} \\ (3x)/(3)\text{ = }(5.544)/(3) \\ x\text{ = 1.848 moles} \end{gathered}`

Hence, the number of moles of aluminum chloride is 1.848 moles

Finally, find the mass of aluminum chloride using the below formula

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

Recall, that the molar mass of aluminum chloride is 133.34 g/mol

`\begin{gathered} 1.848\text{ = }(mass)/(133.34) \\ cross\text{ multiply} \\ mass\text{ = 1.848 }*\text{ 133.34} \\ mass\text{ 246.41 grams} \end{gathered}`

Therefore, the mass of aluminum chloride produced is 246.41 grams