1. The reaction of fluorine with ammonia produces dinitrogen tetrafluoride and hydrofluoric acid.
F2(g) + NH3(g) „³ N2F4(g) + HF(g)
a. If you have 66.5g NH3, how many grams of F2 are required for complete reaction?
b. How many grams of NH3 are required to produce 4.65g HF?
c. How many grams of N2F4 can be produced from 225g F2?
2. Lithium nitride reacts with water to form ammonia and aqueous lithium hydroxide.
Li3N(s) + H2O(l) „³ NH3(g) + LiOH(aq)
a. What mass of water is needed to react with 32.9g Li3N?
b. How many molecules of NH3 are produced if 145g of LiOH are also produced?
how to convert moles to molecules (use Avogadro number):
http://www.ehow.com/how_6557465_convert-moles-molecules.html
http://www.jiskha.com/display.cgi?id=1301167744
a. To determine the grams of F2 required for the reaction, we need to use stoichiometry. From the balanced equation, we can see that the mole ratio between F2 and NH3 is 1:1. We are given the mass of NH3 (66.5g), so we need to convert it to moles using the molar mass of NH3 (17.03 g/mol).
Moles of NH3 = Mass of NH3 / Molar mass of NH3
= 66.5g / 17.03 g/mol
= 3.9 mol NH3
Since the mole ratio between F2 and NH3 is 1:1, the moles of F2 required will be the same as the moles of NH3. Therefore, the number of grams of F2 required for the complete reaction is also 66.5g.
b. To determine the grams of NH3 required to produce 4.65g of HF, we again need to use stoichiometry. From the balanced equation, we can see that the mole ratio between NH3 and HF is 1:1. We are given the mass of HF (4.65g), so we need to convert it to moles using the molar mass of HF (20.01 g/mol).
Moles of HF = Mass of HF / Molar mass of HF
= 4.65g / 20.01 g/mol
= 0.232 mol HF
Since the mole ratio between NH3 and HF is 1:1, the moles of NH3 required will also be 0.232 mol. Therefore, the number of grams of NH3 required to produce 4.65g of HF is 0.232g.
c. To determine the grams of N2F4 that can be produced from 225g of F2, we need to use stoichiometry. From the balanced equation, we can see that the mole ratio between F2 and N2F4 is 1:1. We are given the mass of F2 (225g), so we need to convert it to moles using the molar mass of F2 (38.0 g/mol).
Moles of F2 = Mass of F2 / Molar mass of F2
= 225g / 38.0 g/mol
= 5.92 mol F2
Since the mole ratio between F2 and N2F4 is 1:1, the moles of N2F4 that can be produced will also be 5.92 mol. Therefore, the number of grams of N2F4 that can be produced from 225g of F2 is 5.92g.
a. To determine the mass of water needed to react with 32.9g of Li3N, we need to use stoichiometry. From the balanced equation, we can see that the mole ratio between Li3N and H2O is 1:3. We are given the mass of Li3N (32.9g), so we need to convert it to moles using the molar mass of Li3N (34.83 g/mol).
Moles of Li3N = Mass of Li3N / Molar mass of Li3N
= 32.9g / 34.83 g/mol
= 0.944 mol Li3N
Since the mole ratio between Li3N and H2O is 1:3, the moles of H2O needed will be three times the moles of Li3N.
Moles of H2O = (3 mol H2O / 1 mol Li3N) * (0.944 mol Li3N)
= 2.83 mol H2O
To convert moles of H2O to grams, we use the molar mass of H2O (18.02 g/mol).
Mass of H2O = Moles of H2O * Molar mass of H2O
= 2.83 mol * 18.02 g/mol
= 51.0g
Therefore, 51.0g of water is needed to react with 32.9g of Li3N.
b. To determine the number of molecules of NH3 produced if 145g of LiOH is also produced, we need to use stoichiometry. From the balanced equation, we can see that the mole ratio between LiOH and NH3 is 1:1. We are given the mass of LiOH (145g), so we need to convert it to moles using the molar mass of LiOH (23.95 g/mol).
Moles of LiOH = Mass of LiOH / Molar mass of LiOH
= 145g / 23.95 g/mol
= 6.06 mol LiOH
Since the mole ratio between LiOH and NH3 is 1:1, the moles of NH3 produced will also be 6.06 mol. To convert moles of NH3 to the number of molecules, we use Avogadro's number (6.022x10^23 molecules/mol).
Number of molecules of NH3 = Moles of NH3 * Avogadro's number
= 6.06 mol * 6.022x10^23 molecules/mol
= 3.65x10^24 molecules
Therefore, 3.65x10^24 molecules of NH3 are produced when 145g of LiOH is produced.