Respuesta :
Answer: 169.3 g of [tex]AlCl_3[/tex] will be produced
Explanation:
To calculate the moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}[/tex]
moles of [tex]Al[/tex]
[tex]\text{Number of moles}=\frac{50.0g}{27g/mol}=1.85moles[/tex]
moles of [tex]Cl_2[/tex]
According to ideal gas equation:
[tex]PV=nRT[/tex]
P = pressure of gas = 1 atm (STP)
V = Volume of gas = 42.7 L
n = number of moles = ?
R = gas constant =[tex]0.0821Latm/Kmol[/tex]
T =temperature =[tex]273K[/tex] (at STP)
[tex]n=\frac{PV}{RT}[/tex]
[tex]n=\frac{1atm\times 42.7L}{0.0820 L atm/K mol\times 273K}=1.90moles[/tex]
[tex]2Al+3Cl_2\rightarrow 2AlCl_3[/tex]
According to stoichiometry:
3 moles of [tex]Cl_2[/tex] reacts with = 2 moles of aluminium
Thus 1.90 moles of [tex]Cl_2[/tex]reacts with= [tex]\frac{2}{3}\times 1.90=1.27[/tex] moles of aluminium
Thus [tex]Cl_2[/tex] is the limiting reagent as it limits the formation of product.
As 3 moles of [tex]Cl_2[/tex] give = 2 moles of [tex]AlCl_3[/tex]
Thus 1.90 moles of [tex]Cl_2[/tex] give =[tex]\frac{2}{3}\times 1.90=1.27moles[/tex] of [tex]AlCl_3[/tex]
Mass of [tex]AlCl_3=moles\times {\text {Molar mass}}=1.27moles\times 133.34g/mol=169.3g[/tex]
Thus 169.3 g of [tex]AlCl_3[/tex] will be produced from the given masses of both reactants.
