Considering the equation:
Li(s)+N₂(g)—->Li₃N(s)
Balancing this equation will give :
6Li(s)+N₂(g)—->2Li₃N(s)
It can be seen that 6 mol of Li reacts with 1 mol of N₂to produce 2 mol of Li₃N.
Here 33.6524g of nitrogen gas reacts with 58.7032g lithium to produce Li₃N.
Moles of Li = 58.7032 g * [tex] \frac{1 mol }{6.94 g} = 8.46 mol Li [/tex]
Moles of [tex] N_{2} = 33.6524 g * \frac{1 mol}{28 g} = 1.2 mol N_{2} [/tex]
Mass of [tex] Li_{3} N from Li = 8.46 mol Li * \frac{2 mol Li_{3}N}{6 mol Li}* \frac{34.83 g Li_{3}N}{1 mol Li_{3}N} = 98.2 g [/tex]
Mass of [tex] Li_{3} N from [tex] N_{2} [/tex] = 1.2 mol [tex] N_{2} [/tex] * \frac{2 mol Li_{3}N}{1 mol N2}* \frac{34.83 g Li_{3}N}{1 mol Li_{3}N} = 98.2 g [/tex] = 83.6 g
