Answer: B. you will need 53.2 g Cl2 for complete reaction and will produce 66.7 g of AlCl3.
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.
To calculate the moles, we use the equation:
[tex]\text{Number of moles of aluminium}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{13.5g}{27g/mol}=0.5moles[/tex]
The balanced reaction is:
[tex]2Al+3Cl_2(g)\rightarrow 2AlCl_3[/tex]
2 moles of aluminium react with= 3 moles of chlorine
Thus 0.5 moles of aluminium react with=[tex]\frac{3}{2}\times 0.5=0.75[/tex] moles of chlorine
Mass of chlorine=[tex]moles\times {\text{Molar Mass}}=0.75\times 71=53.2g[/tex]
2 moles of aluminium produce = 2 moles of aluminium chloride
Thus 0.5 moles of aluminium react with=[tex]\frac{2}{2}\times 0.5=0.5[/tex] moles of aluminium chloride
Mass of aluminium chloride=[tex]moles\times {\text{Molar Mass}}=0.5\times 133.34=66.7g[/tex]
Thus 53.2 g of chlorine is used and 66.7 g of aluminium chloride is produced.
