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How many grams of lithium are required to completely react with 58.5 mL of N2 gas...

How many grams of lithium are required to completely react with 58.5 mL of N2 gas measured at STP?
6 Li(s) + N2(g) = 2 Li3N(s)

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Answer #1

Answer:- At STP, 22.4 L of gas has 1 mol value. So at STP, the number of moles in 58.5 ml is-

n_{N_{2}}=\frac{58.5/1000}{22.4}\ mol

\Rightarrow n_{N_{2}}=0.002611\ mol

from the equation, we see that 1 mol of N2 needs 6 mol of Lithium. So for the above mol of Nitrogen, we need-

\Rightarrow n_{Li}=0.002611*6\ mol=0.01566\ mol

Hence the number of grams of Lithium required is-

W_{Li}=0.01566*molar\ mass\ of\ Lithium

\Rightarrow W_{Li}=0.01566*6.94\ gram

\boldsymbol{\Rightarrow W_{Li}=0.1087\ gram}

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