Lithium nitride reacts with water to form ammonia and aqueous lithium hydroxide.
Li,N(s) + 3H,0(L) - > NH(g) + 3LiOH(aq)
a. What mass of water is needed to react with 32.9 g LigN?
b. When the above reaction takes place, how many molecules of NH, are produced?
c. Calculate the number of grams of LigN that must be added to an excess of water to produce 15.0 L NH (at STP).
a. To determine the mass of water needed to react with 32.9 g of Li3N, we first need to balance the equation:
Li3N + 3H2O -> NH3 + 3LiOH
From the balanced equation, we see that 1 mole of Li3N reacts with 3 moles of H2O. Now we can calculate the molar mass of Li3N:
3Li = 3 x 6.94 g/mol = 20.82 g
N = 14.01 g/mol
Total molar mass = 20.82 + 14.01 x 3 = 62.85 g/mol
Now, we can calculate the number of moles of Li3N in 32.9 g:
32.9 g Li3N / 62.85 g/mol Li3N = 0.523 mol Li3N
Since 1 mole of Li3N reacts with 3 moles of H2O, we need 3 x 0.523 mol = 1.569 mol H2O to react with 32.9 g of Li3N. Finally, we can calculate the mass of water needed:
1.569 mol H2O x 18.015 g/mol H2O = 28.26 g H2O
Therefore, 28.26 g of water is needed to react with 32.9 g of Li3N.
b. To determine the number of molecules of NH3 produced, we need to use the ideal gas law equation:
PV = nRT
At STP (standard temperature and pressure), 1 mole of gas occupies 22.4 L. Therefore, in 15.0 L of NH3, there are:
15.0 L NH3 / 22.4 L/mol = 0.6705 mol NH3
Since 1 mole of NH3 contains 6.022 x 10^23 molecules, the number of molecules of NH3 produced is:
0.6705 mol NH3 x 6.022 x 10^23 molecules/mol = 4.03 x 10^23 molecules NH3
c. To calculate the number of grams of Li3N needed to produce 15.0 L of NH3 at STP, we first need to determine the number of moles of NH3:
15.0 L NH3 / 22.4 L/mol = 0.6705 mol NH3
From the balanced equation, we know that 1 mole of Li3N reacts with 1 mole of NH3. Therefore, we need 0.6705 mol of Li3N. Finally, we can calculate the mass of Li3N needed:
0.6705 mol Li3N x 62.85 g/mol Li3N = 42.16 g Li3N
Therefore, 42.16 grams of Li3N must be added to an excess of water to produce 15.0 L of NH3 at STP.