Question

When Al metal reacts with HBr, aluminum bromide and hydrogen gas are formed. If you started with 41.2 grams of HBr, how many grams of aluminum bromide would you expect to form?

Answer 1

Molar mass of HBr,

MM = 1*MM(H) + 1*MM(Br)

= 1*1.008 + 1*79.9

= 80.908 g/mol

mass of HBr = 41.2 g

mol of HBr = (mass)/(molar mass)

= 41.2/80.91

= 0.5092 mol

Balanced chemical equation is:

2 Al + 6 HBr —> 2 AlBr3 + 3 H2

According to balanced equation

mol of AlBr3 formed = (2/6)* moles of HBr

= (2/6)*0.5092

= 0.1697 mol

Molar mass of AlBr3,

MM = 1*MM(Al) + 3*MM(Br)

= 1*26.98 + 3*79.9

= 266.68 g/mol

mass of AlBr3 = number of mol * molar mass

= 0.1697*2.667*10^2

= 45.27 g

Answer: 45.3 g