If water I'd added to magnesium nitride, ammonia gas is produced when the mixture is heated
Mg3N2 (s) + 3H2O (l) ----> 3MgO) (s) + 2NH) (g)
*Only the numbers at the beginning aren't subscripts*
If 10.3 g of magnesium nitride is treated with water, what volume of ammonia gas would be collected at 24 degrees C and 752 mm Hg?
I don't know how to set this up D:
Here is a solved example stoichiometry problem. Just follow the steps. For the volume, use PV = nRT.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To determine the volume of ammonia gas produced, you need to 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 calculate the number of moles of ammonia gas produced:
1) Convert the given mass of magnesium nitride (Mg3N2) to moles:
Molar mass of Mg3N2 = (3 × molar mass of Mg) + (2 × molar mass of N)
Molar mass of Mg = 24.31 g/mol
Molar mass of N = 14.01 g/mol
Molar mass of Mg3N2 = (3 × 24.31 g/mol) + (2 × 14.01 g/mol) = 100.94 g/mol
Number of moles of Mg3N2 = mass of Mg3N2 / molar mass of Mg3N2
= 10.3 g / 100.94 g/mol
2) Calculate the number of moles of ammonia gas produced:
According to the balanced chemical equation, 1 mole of Mg3N2 reacts to produce 2 moles of NH3.
Therefore, moles of NH3 = 2 × moles of Mg3N2
Now that we have the number of moles of ammonia gas produced, we can calculate the volume of gas using the ideal gas law equation.
3) Convert the given temperature from degrees Celsius to Kelvin:
T(K) = T(°C) + 273.15
T = 24°C + 273.15
4) Convert the given pressure from mm Hg to atm:
P(atm) = P(mm Hg) / 760
Now we have all the necessary values to plug into the ideal gas law equation.
PV = nRT
(Volume in liters) × (pressure in atm) = (moles of NH3) × (0.0821 L·atm/(mol·K)) × (temperature in K)
Substitute the given values:
(Volume in liters) × (752 mm Hg / 760) = (2 moles × 0.0821 L·atm/(mol·K) × (24°C + 273.15 K)
Simplify and solve for volume:
Volume in liters = (2 moles × 0.0821 L·atm/(mol·K) × (24°C + 273.15 K)) / (752 mm Hg / 760)
Calculate the volume using the given values and solve for the volume of ammonia gas produced.
To solve this problem, you need to use the given balanced chemical equation and utilize the concept of stoichiometry. Stoichiometry is a relationship between the quantities of reactants and products in a chemical reaction.
First, let's convert the mass of magnesium nitride (Mg3N2) into moles. To do this, you'll need to know the molar mass of the compound. The molar mass of magnesium (Mg) is 24.31 g/mol, and the molar mass of nitrogen (N) is 14.01 g/mol. Since there are three magnesium atoms and two nitrogen atoms in a formula unit of Mg3N2, the molar mass of magnesium nitride is 101.10 g/mol.
Therefore, for 10.3 g of Mg3N2, we can calculate the number of moles:
moles of Mg3N2 = mass / molar mass
moles of Mg3N2 = 10.3 g / 101.10 g/mol
Next, we need to use the stoichiometry to calculate the number of moles of ammonia gas (NH3) produced. According to the balanced chemical equation, for every 2 moles of NH3 produced, we need 1 mole of Mg3N2. This means the mole ratio is 1:2.
moles of NH3 = moles of Mg3N2 * (2 moles of NH3 / 1 mole of Mg3N2)
Now that we have the number of moles of ammonia gas, we can use the ideal gas law to find the volume in liters. The ideal gas law equation is:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature (in Kelvin)
First, convert the given temperature from degrees Celsius to Kelvin by adding 273.15:
Temperature (in Kelvin) = 24°C + 273.15 = 297.15 K
Now, plug in the values into the ideal gas law equation:
V = (nRT) / P
V = (moles of NH3 * R * T) / P
Finally, substitute the given values into the equation:
V = (moles of NH3 * 0.0821 L.atm/mol.K * 297.15 K) / 0.988 atm (since 752 mm Hg is equivalent to 0.988 atm)
Calculate the volume (V) to find the final answer.