Question

2Al + 3Cl2 --> 2AlCl3If 15.2 g of aluminum reacts with 39.1g of chlorine, how many g of AlCl3 forms? (2 decimal places)

Answer 1

The first step to solve this question is to convert the given masses to moles using the corresponding molecular weight:

`15.2gAl\cdot(molAl)/(27gAl)=0.56molAl`
`39.1gCl\cdot(molCl_2)/(35.45gCl_2)=1.1molCl_2`

Divide each of the results by the coefficient of each reactant in the equation. It means, divide the amount of Al by 2 and the amount of Cl2 by 3:

`\begin{gathered} (0.56molAl)/(2molAl)=0.28 \\ (1.1molCl_2)/(3molCl_2)=0.367 \end{gathered}`

It means that the limiting reactant is Aluminium and we have to base our calculations on this reactant. From the equation we know than 2 moles of Al produce 2 moles of AlCl3.

`0.56molAl\cdot(2molAlCl_3)/(2molAl)=0.56molAlCl_3`

Now, convert this amount of AlCl3 to mass using its molecular weight:

`0.56molAlCl_3\cdot(133.33gAlCl_3)/(molAlCl_3)=74.66gAlCl_3`

It means that the mass of AlCl3 formed is 74.66g.