Question

lithium reacts with nitrogen gas according to the following reaction:
6li(s) + n₂(g)→2li₃n(s)
what mass of lithium is required to react completely with 57.3 ml of n₂ gas at stp?
express your answer in grams.
m = g

Explanation:

Step1: Calculate the number of moles of \(N_2\) gas

At STP (Standard Temperature and Pressure), \(1\) mole of any ideal gas occupies \(22.4\) L.
First, convert the volume of \(N_2\) from mL to L: \(V = 57.3\space mL=57.3\times10^{- 3}\space L\)
Using the formula \(n=\frac{V}{V_m}\) (where \(V_m = 22.4\space L/mol\)), we have \(n_{N_2}=\frac{57.3\times 10^{-3}\space L}{22.4\space L/mol}\)
\(n_{N_2}\approx2.56\times10^{-3}\space mol\)

Step2: Use the mole ratio from the balanced chemical equation

From the balanced equation \(6Li(s)+N_2(g)\to2Li_3N(s)\), the mole ratio of \(Li\) to \(N_2\) is \(n_{Li}:n_{N_2}=6:1\)
So \(n_{Li}=6\times n_{N_2}\)
Substitute \(n_{N_2} = 2.56\times 10^{-3}\space mol\) into the equation: \(n_{Li}=6\times2.56\times10^{-3}\space mol = 1.54\times10^{-2}\space mol\)

Step3: Calculate the mass of \(Li\)

The molar mass of \(Li\) is \(M_{Li}=6.94\space g/mol\)
Using the formula \(m = n\times M\), we have \(m_{Li}=n_{Li}\times M_{Li}\)
Substitute \(n_{Li}=1.54\times 10^{-2}\space mol\) and \(M_{Li} = 6.94\space g/mol\)
\(m_{Li}=1.54\times10^{-2}\space mol\times6.94\space g/mol\approx0.107\space g\)

Answer:

\(0.107\space g\)