Question

What is the limiting reactant when 1.50 g of lithium and 1.50 g of nitrogen combine to form lithium nitride, a component of advanced batteries, according to the following unbalanced equation?\(Li + {N_2} \to L{i_3}N\).

Short answer

Li is the limiting reactant.

Step-by-step solution

Balanced chemical equation

Balance the chemical equation.

\(3Li + \frac{1}{2}{N_2} \to L{i_3}N\)

Multiply both sides by 2.

\(6Li + {N_2} \to 2L{i_3}N\)

Determine the molar mass

Find the molar mass.

1 mol Li=6.941g

1 mol \({N_2} = 2\left( {14.007} \right) = 28.014g\).

Determine limiting reactant

Find the mass of \({N_2}\)need to react completely with 1.59g Li.

\(mol\,{H_3}P{O_4} = 0.50\,mol\,Cr\left( {\frac{{2\,mol\,{H_3}P{O_4}}}{{2\,mol\,Cr}}} \right) = 0.50\,mol\,{H_3}P{O_4}\)

Mass\({N_2}\)=\(1.50gLi\left( {\frac{{1molLi}}{{6.941gLi}}} \right)\left( {\frac{{1mol{N_2}}}{{6molLi}}} \right)\left( {\frac{{28.014g{N_2}}}{{1mol{N_2}}}} \right) = 1.01g{N_2}\)

1.50 g\({N_2}\)> 1.01 g\({N_2}\)

As N₂ is in excess quantity it can be concluded Li is the limiting reactant