Question

aluminum nitrite and ammonium cloride react to form aluminium chloride. nitrogen and water. what mass of alluminum chloride is present of 43g of ammonium nitrite have reacted completely

Answer 1

Answer:

35.7 g of NH₄Cl

Explanation:

The reaction is:

Al(NO₂)₃ +  3NH₄Cl → AlCl₃ + 3N₂ + 6H₂O

Ratio is 3:1. 3 mol of ammoium chloride react to produce 1 mol of chloride

We convert the mass of ammoium chloride to moles:

43 g . 1 mol/ 53.45 g = 0.804 moles

(0.804 . 1) / 3 = 0.268 moles of AlCl₃ are formed

Let's convert the moles to mass → 0.268 mol . 133.33 g/1mol = 35.7 g