Respuesta :
Answer:
1) 4.0744 grams of aluminum chloride can be obtained.
2) 0.7652 grams of chlorine that react with 0.194 g of aluminum.
Explanation:
[tex] 2Al(s) + 3Cl_2(g)\rightarrow 2AICI_3[/tex]
1) Moles of aluminum : [tex]\frac{0.824 g}{27 g/mol}=0.03052 mol[/tex]
According to reaction, 2 moles of aluminum gives 2 moles of aluminium chloride.
Then 0.03052 mol of aluminum will give with :
[tex]\frac{2}{2}\times 0.03052 mol=0.03052 mol[/tex] of aluminium chloride.
Mass of 0.03052 moles of aluminium chloride :
[tex]0.03052 mol\times 133.5 g/mol=4.0744 g[/tex]
4.0744 grams of aluminum chloride can be obtained.
2) Moles of Aluminium : [tex]\frac{0.194 g}{27 g/mol}=0.007185 mol[/tex]
According to reaction, 2 moles of aluminum reacts with 3 moles of chlorine gas.
Then 0.007185 mol of aluminum will react with:
[tex]\frac{3}{2}\times 0.007185 mol=0.01077 mol[/tex] of chlorine gas.
Mass of 0.01077 mol moles of chlorine gas:
[tex]0.01077 mol\times 71 g/mol=0.7652 g[/tex]
0.7652 grams of chlorine that react with 0.194 g of aluminum.
