Lithium nitride reacts with water to form ammonia and aqueous lithium hydroxide.
Li3N(s) + 3H2O(l) ==> NH3(g) + 3LiOH(aq)
a)What mass of water is needed to react with 32.9g Li3N?
b)When the above reaction takes place, how many molecules of NH3 are produced?
c)Calculate the number of grams of Li3N that must be added to an excess of water to produce 15.0L NH3(at STP).
For a. I got 51.03 grams of H2O
I dont understand and I honestly don't get chemistry at all but I'm trying
Lithium nitride reacts with water to form ammonia and aqueous lithium hydroxide.
Li3N(s) + 3H2O(l) ==> NH3(g) + 3LiOH(aq)
a)What mass of water is needed to react with 32.9g Li3N?
b)When the above reaction takes place, how many molecules of NH3 are produced?
c)Calculate the number of grams of Li3N that must be added to an excess of water to produce 15.0L NH3(at STP).
Kim
Apr 5, 2010
1. Write and balance the equation. You have that.
a. Convert 32.9 g Li3N to moles. moles = grams/molar mass.
Using the coefficients in the balanced equation, convert moles Li3N to moles H2O.
Then convert moles H2O to grams. g = moles x molar mass.
b. To obtain moles NH3, use the same process as in part a. Then remember that 1 mole of any material contains 6.022 x 10^23 molecules.
c. Same procedure as in a except convert 15.0 L NH3 at STP (remember 1 mole of a gas occupies 22.4 L at STP) to moles, convert moles NH3 to moles Li3N, then grams = moles x molar mass.
Post your work if you get stuck.
DrBob222
Apr 5, 2010
For a I have .948 mol Li3N and for h20 I have 17.07848544 grams so now what do I do
Abby
Apr 20, 2012
I got answers for a and b but I don't know where to begin on c
Abby
Apr 20, 2012
I dont understand and I honestly don't get chemistry at all but I'm trying
Anonymous
Apr 23, 2021
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levi mccarthey
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To answer these questions, we need to determine the stoichiometry of the reaction, which is the ratio of moles between the reactants and products.
a) To find the mass of water needed to react with 32.9g of Li3N, we need to calculate the molar mass of Li3N and use stoichiometry.
1. Calculate the molar mass of Li3N:
Li: 6.941 g/mol
N: 14.007 g/mol
3Li + N = 3(6.941) + 14.007 = 34.264 g/mol
2. Convert 32.9g of Li3N to moles:
32.9g / 34.264 g/mol = 0.959 mol of Li3N
3. Use stoichiometry to find the moles of water needed:
From the balanced equation, the stoichiometric ratio of Li3N to H2O is 1:3.
So, moles of H2O = (0.959 mol of Li3N) x (3 mol of H2O / 1 mol of Li3N) = 2.877 mol of H2O
4. Convert moles of water to grams:
Molar mass of H2O = 18.015 g/mol
Mass of water = (2.877 mol of H2O) x (18.015 g/mol) = 51.822 g of H2O
Therefore, 51.822 grams of water are needed to react with 32.9 grams of Li3N.
b) To determine the number of ammonia (NH3) molecules produced, we need to use Avogadro's number and the stoichiometry of the reaction.
1. From the balanced equation, we can see that one mole of Li3N reacts to produce one mole of NH3.
So, the number of moles of NH3 produced is equal to the number of moles of Li3N.
2. Convert moles of Li3N to moles of NH3:
Number of moles of NH3 = 0.959 mol (calculated in part a)
3. Use Avogadro's number (6.022 x 10^23) to find the number of NH3 molecules:
Number of NH3 molecules = (0.959 mol) x (6.022 x 10^23 molecules/mol)
Therefore, the number of NH3 molecules produced is approximately 5.78 x 10^23.
c) To find the number of grams of Li3N that must be added to produce 15.0L of NH3 at STP, we need to use the ideal gas law and stoichiometry.
1. Convert 15.0L of NH3 to moles using the ideal gas law:
PV = nRT (where P is pressure, V is volume, n is moles, R is the ideal gas constant, and T is temperature in Kelvin)
At STP (Standard Temperature and Pressure), P = 1 atm and T = 273 K.
So, n = PV / RT = (1 atm) x (15.0 L) / (0.0821 L·atm/mol·K) x (273 K) ≈ 0.750 mol of NH3
2. Use stoichiometry to find the moles of Li3N:
From the balanced equation, the stoichiometric ratio of Li3N to NH3 is 1:1.
Therefore, the moles of Li3N needed is also 0.750 mol.
3. Convert moles of Li3N to grams:
Molar mass of Li3N = 34.264 g/mol
Mass of Li3N = 0.750 mol x 34.264 g/mol ≈ 25.698 g of Li3N
Therefore, approximately 25.698 grams of Li3N must be added to excess water to produce 15.0L of NH3 at STP.
For a I have .948 mol Li3N and for h20 I have 17.07848544 grams so now what do I do
I got answers for a and b but I don't know where to begin on c
1. Write and balance the equation. You have that.
a. Convert 32.9 g Li3N to moles. moles = grams/molar mass.
Using the coefficients in the balanced equation, convert moles Li3N to moles H2O.
Then convert moles H2O to grams. g = moles x molar mass.
b. To obtain moles NH3, use the same process as in part a. Then remember that 1 mole of any material contains 6.022 x 10^23 molecules.
c. Same procedure as in a except convert 15.0 L NH3 at STP (remember 1 mole of a gas occupies 22.4 L at STP) to moles, convert moles NH3 to moles Li3N, then grams = moles x molar mass.
Post your work if you get stuck.