If water is added to magnesium nitride, ammonia gas is produced when the mixture is heated.
Mg3N2(s) + 3 H2O(l) => 3 MgO(s) + 2 NH3(g)
If 11.1 g of magnesium nitride is treated with water, what volume of ammonia gas would be collected at 23°C and 739 mm
mols Mg3N2 = grams/molar mass
Convert mols Mg3N2 to mols NH3 using the coefficients in the balanced equation.
Use PV = nRT to convert mols NH3 to V(in L) at the conditions listed.
Note:This answer is based upon no excess H2O being present since NH3 gas is quite soluble in H2O.
To find the volume of ammonia gas produced, we need to use the ideal gas law equation, which is given by:
PV = nRT
Where:
P = pressure of the gas (in atm)
V = volume of the gas (in liters)
n = number of moles of the gas
R = ideal gas constant (0.0821 L.atm/(mol.K))
T = temperature of the gas (in Kelvin)
First, we need to calculate the number of moles of ammonia gas produced.
Step 1: Convert the mass of magnesium nitride to moles
Molar mass of Mg3N2 = (3*24.31 g/mol) + (2*14.01 g/mol)
= 100.95 g/mol
Number of moles of Mg3N2 = Mass / Molar mass
= 11.1 g / 100.95 g/mol
= 0.1099 mol
Step 2: Using the stoichiometry of the balanced equation, we can determine the moles of NH3 produced.
From the balanced equation: 2 moles of NH3 is produced per 1 mole of Mg3N2.
Number of moles of NH3 = 2 * 0.1099 mol
= 0.2198 mol
Step 3: Convert the temperature to Kelvin
Temperature in Kelvin = 23°C + 273.15
= 296.15 K
Step 4: Plug the values into the ideal gas law equation
PV = nRT
(739 mm Hg) * (1 atm/760 mm Hg) = (0.2198 mol) * (0.0821 L.atm/(mol.K)) * (296.15 K)
Step 5: Solve for V (volume of ammonia gas)
V = (0.2198 mol) * (0.0821 L.atm/(mol.K)) * (296.15 K) / (1 atm/760 mm Hg)
≈ 4.72 L
Therefore, approximately 4.72 liters of ammonia gas would be collected at 23°C and 739 mm Hg.
To solve this problem, we need to use the stoichiometry of the balanced chemical equation provided. Stoichiometry is the calculation of quantities in chemical reactions based on the given amount of reactants and products.
First, we need to determine the number of moles of magnesium nitride (Mg3N2) that we have. To do this, we need to convert the given mass (11.1 g) into moles using the molar mass of magnesium nitride.
The molar mass of Mg3N2 is calculated by adding up the atomic masses of magnesium (Mg) and nitrogen (N):
Mg: 24.31 g/mol
N: 14.01 g/mol
Molar mass of Mg3N2 = (3 * Mg) + (2 * N)
= (3 * 24.31 g/mol) + (2 * 14.01 g/mol)
= 100.93 g/mol
Now, we can calculate the number of moles of Mg3N2:
Number of moles = Given mass / Molar mass
= 11.1 g / 100.93 g/mol
= 0.11 mol
From the balanced chemical equation, we see that the ratio of Mg3N2 to NH3 is 1:2. Therefore, the number of moles of NH3 produced will be twice the number of moles of Mg3N2 used.
Number of moles of NH3 = 2 * Number of moles of Mg3N2
= 2 * 0.11 mol
= 0.22 mol
Now, we need to use the ideal gas law to calculate the volume of ammonia gas produced. The ideal gas law states:
PV = nRT
where P is the pressure in atm, V is the volume in liters, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin.
First, let's convert the given temperature of 23°C to Kelvin:
T(K) = T(°C) + 273.15
= 23°C + 273.15
= 296.15 K
Next, we need to convert the given pressure of 739 mmHg to atm:
P(atm) = P(mmHg) / 760
= 739 mmHg / 760
= 0.972 atm
Now, we can calculate the volume of NH3 gas produced using the ideal gas law:
V = nRT / P
= (0.22 mol) * (0.0821 L·atm/mol·K) * (296.15 K) / (0.972 atm)
≈ 6.59 L
Therefore, approximately 6.59 liters of ammonia gas would be collected at 23°C and 739 mmHg.