Question

Using the reaction 6Li(s) + N2 --> 2Li3N(s), how many grams of Lithium are needed to completely react with 67.0 mL of N2 gas at STP?

Answer 1

because things react in terms of moles (i.e. molecules) you need to convert what you're given to moles.

SO convert 67 mL of \(N_2\) to moles.

then set up a ratio with the stoichiometric coefficients (like i posted before)

then solve for moles of Li, then convert those to grams.

Answer 2

ps. to find the moles of \(N_2\) you can use the ideal gas law, or simply remember that at STP, 1 mole of any gas = 22.414 L

Answer 3

I Know 67.0 mL of N2 is 2.3928 moles but once I get there the conversion factors get me stuck
\[67.0 g N_2*\frac{ 1 mol N_2 }{ 28.0 g/mol N_2}= 2.398\]
but from there I am unclear as to which coefficients I should use

Answer 4

hm you didn't convert mL to L, 0.067 L

\(\dfrac{0.067L}{22.414L}=0.0029892031765861~moles\)

then you set up a ratio using this equation

6Li(s) + N2 --> 2Li3N(s)

so the ratio looks like this: \(\dfrac{n_{Li}}{6}=\dfrac{n_{N_2}}{1}\)

Answer 5

and then you just cross multiply?

Answer 6

I would assume so.

Answer 7

yep, solve for \(n_{Li}\), then convert to grams.