Lithium nitride reacts with water to form ammonia and aqueous lithium hydroxide.

Question

Li,N(s) + 3H,0(L) - > NH(g) + 3LiOH(aq)
a. What mass of water is needed to react with 32.9 g Li3N?
b. When the above reaction takes place, how many molecules of NH3, are produced?
c. Calculate the number of grams of Li3Nthat must be added to an excess of water to produce 15.0 L NH3 (at STP).

Answer 1

a. First, we need to determine the molar mass of Li3N:
Li = 6.94 g/mol
N = 14.01 g/mol
Molar mass of Li3N = 3(6.94) + 14.01 = 34.83 g/mol

Next, we need to calculate the number of moles of Li3N present in 32.9 g:
moles = mass / molar mass = 32.9 g / 34.83 g/mol = 0.944 mol

Since the reaction requires 3 moles of water for every mole of Li3N, we need 3 moles of water for every 0.944 moles of Li3N:
moles of water = 3 x 0.944 = 2.832 mol

Now, we need to convert moles of water to grams:
mass = moles x molar mass of water = 2.832 mol x 18.02 g/mol = 51.00 g

Therefore, 51.00 grams of water is needed to react with 32.9 g of Li3N.

b. From the balanced equation, we see that 1 mole of Li3N produces 1 mole of NH3. Therefore, 0.944 moles of Li3N will produce 0.944 moles of NH3, which is equivalent to 0.944 x 6.022 x 10^23 molecules of NH3. This gives us 5.68 x 10^23 molecules of NH3.

c. First, we need to calculate the number of moles of NH3 that will be produced:
From the ideal gas law, we can calculate the number of moles of NH3:
PV = nRT
n = PV/RT
n = (1 atm)(15.0 L) / (0.0821 L.atm/mol.K)(273 K)
n = 0.681 mol

Since 1 mole of Li3N produces 1 mole of NH3, we need 0.681 moles of Li3N to produce 15.0 L of NH3 at STP.

Now, we need to convert moles of Li3N to grams:
mass = moles x molar mass = 0.681 mol x 34.83 g/mol = 23.77 g

Therefore, 23.77 grams of Li3N must be added to an excess of water to produce 15.0 L of NH3 at STP.