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
If 5.00 g Br2 and 1.10 g NH3 react according to the equation below, what is the maximum mass of...
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please explain for me what each step do , thank you
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