Answer to Question #120302 in General Chemistry for bob

Question

Aluminum nitrite and ammonium chloride react to form aluminum chloride, nitrogen, and water. What mass of AlCl3 is present after 99.75 g aluminum nitrite and 84.94 g ammonium chloride react completely?

Answer 1

Al(NO)₃+3NH₄Cl=AlCl₃+3N₂+6H₂0

$n(Al(NO)3)=\frac{m}{M}=\frac{99.75}{164.9965}=0.60$

$n(NH4Cl)=\frac{m}{M}=\frac{84.94}{53.4920}=1.59$

aluminum nitrite is in short supply, then we calculate it

n(AlCl₃)=0.60

$m=n\times M=0.6\times 133.3405=80.00$ g