Question

Please help me to find yield of Alcl3 in gram

Answer 1

Theoretical yield is calculated based on the stoichiometry of the chemical equation. First, we have to find the number of moles of each reactant using the molar mass of each of them. You can find the molar mass using the periodic table: the molar mass of aluminum (Al) is 26.98 g/mol and the molar mass of chlorine gas (Cl2) is 70.8 g/mol:

`34.0\text{ g Al}\cdot\frac{1\text{ mol Al}}{26.98\text{ g Al}}=1.26\text{ moles Al.}`
`39.0gCl_2\cdot\frac{1\text{ mol }Cl_2}{70.8\text{ g }Cl_2}=0.551\text{ moles }Cl_2.`

The next step is to find how many moles of AlCl3 are produced by each reactant. You can see that in the chemical equation, 2 moles of Al produces 2 moles of AlCl3:

`1.26\text{ moles Al}\cdot\frac{2molesAlCl_3}{2\text{ moles Al}}=1.26molesAlCl_3.`

And 3 moles of Cl2 produces 2 moles of AlCl3:

`0.551molesCl_2\cdot(2molesAlCl_3)/(3molesCl_2)=0.367molesAlCl_3.`

You can realize that the maximum, amount of moles that are produced of AlCl3 is 0.367 moles because Cl2 would be the limiting reactant and there is an excess of Al. For this value, we have to find its mass using the molar mass of AlCl3 which is 133.2 g/mol:

`\text{0}.367molesAlCl_3\cdot(133.2gAlCl_3)/(1molAlCl_3)=48.88gAlCl_3.`

The answer would be that the theoretical yield of AlCl3 is 48.88 g.