Magnesium burns in air to produce magnesium oxide, MgO, and magnesium nitride, Mg3N2. Magnesium nitride reacts with water to give ammonia.
Mg3N2(s) + 6 H2O(l) → 3 Mg(OH)2(s) + 2 NH3(g)
What volume of ammonia gas at 24°C and 753 mmHg will be produced from 4.46 g of magnesium nitride?
A stoichiometry problem. Just follow the steps in this example.
http://www.jiskha.com/science/chemistry/stoichiometry.html
explain how the mass of magnesium changes using a particle modle
This is a question; laughing gas is prepared by heating a mixture if sodium trioxonitrate v and ammonium tetraoxosulphate vi what happens when a candle is placed in a gas jar of this gas
To solve this problem, we will use the Ideal Gas Law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = moles of gas
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature (in Kelvin)
First, let's determine the number of moles of magnesium nitride (Mg3N2) using its molar mass.
The molar mass of Mg3N2 can be calculated as:
Mg: 24.31 g/mol (atomic mass of magnesium)
N: 14.01 g/mol (atomic mass of nitrogen)
Molar Mass of Mg3N2 = 3 * Atomic Mass of Magnesium + 2 * Atomic Mass of Nitrogen
= (3 * 24.31 g/mol) + (2 * 14.01 g/mol)
= 100.93 g/mol
Now, we can calculate the moles of Mg3N2:
Moles of Mg3N2 = Mass of Mg3N2 / Molar Mass of Mg3N2
= 4.46 g / 100.93 g/mol
= 0.0442 mol
Next, we need to convert the temperature from Celsius to Kelvin. The formula for conversion is:
T(K) = T(°C) + 273.15
T(K) = 24°C + 273.15 = 297.15 K
Now, we can use the ideal gas law equation to determine the volume of ammonia gas (NH3):
PV = nRT
V = (n * R * T) / P
V = (0.0442 mol * 0.0821 L.atm/mol.K * 297.15 K) / (753 mmHg / 760 mmHg/atm)
Finally, we can calculate the volume of ammonia gas:
V = 0.0175 L
Therefore, the volume of ammonia gas produced from 4.46 g of magnesium nitride at 24°C and 753 mmHg is approximately 0.0175 Liters.