Question

If 5.00 g Br2 and 1.10 g NH3 react according to the equation below, what is the maximum mass of ammonium bromide produced?
____ Br2( ) + ____ NH3(g) ____ NH4Br(s) + ____ N2(g)

Answer 1

The balanced equation is:

3 Br2(l) + 8 NH3(g) => 6 NH4Br(s) + 1 N2(g)

Theoretical moles of Br2 : NH3 = 3 : 8

Moles of Br2 = mass/molar mass of Br2

= 5.00/159.81 = 0.03129 mol

Moles of NH3 = mass/molar mass of NH3

= 1.10/17.03 = 0.06459 mol

Experimental moles of Br2 : NH3 = 0.03129 : 0.06459

= 0.4844 : 1 = 3.875 : 8

Since Br2 is in excess, NH3 is the limiting reactant

Moles of NH4Br = 6/8 x moles of NH3

= 6/8 x 0.06459 = 0.04844 mol

Mass of NH4Br = moles x molar mass of NH4Br

= 0.04844 x 97.94

= 4.74 g