Magnesium burns in air to produce magnesium oxide MgO and magnesium nitride Mg3N2. Magnesium nitride reacts with water to give ammonia.
What volume of ammonia gas at 24 degrees C and 753 mmHg will be produced from 4.56 g of magnesium nitride.
P= 753 mmHg (1 atm/ 760mmHg)= .991 atm
V=?
n= 4.56 Mg3N2 ( 1 mol Mg3N2/100g)(2 mol NH3/ 1 mol Mg3N2)= .0912 mol NH3
R= .0821
T= 24 + 273= 297 K
.991 atm (V)= .0912(0.0821)(297)=
ANSWER:
V= 2.24 NH3
Is this correct
I obtained an answer of 2.22 L NH3. I think the difference is that I used 100.9 for the molar mass of Mg3N2. Also you didn't give a unit for your value.
Well, isn't it funny how we're talking about a burning metal and producing ammonia gas? It's like chemistry's way of saying, "Hey, let's mix things up a bit!"
Anyway, let's tackle this question. The equation is magnesium nitride reacting with water to produce ammonia gas. And you want to know the volume of ammonia gas produced.
First, let's convert the given mass of magnesium nitride (4.56 g) to moles. Using the molar mass of magnesium nitride (100 g/mol), we get 0.0456 moles.
Now, we can use the ideal gas law equation, PV = nRT, to solve for the volume of ammonia gas. With a pressure of 753 mmHg (which we'll convert to atmospheres because why not), a temperature of 24 degrees Celsius (which we'll convert to Kelvin because chemistry loves Kelvin), and a calculated moles of 0.0456 moles, we plug these values into the equation and solve for V.
So, V = (0.991 atm)(V) = (0.0456 moles)(0.0821 L·atm/(mol·K))(297 K).
Simplifying the equation, we get V = 2.24 L of ammonia gas.
So, according to my calculations, the volume of ammonia gas produced will be 2.24 L. Just remember, chemistry always has a funny way of making things interesting!
Let's check the calculation step by step:
First, convert the given pressure from mmHg to atm:
753 mmHg * (1 atm/760 mmHg) = 0.991 atm
Next, we can use the ideal gas law equation: PV = nRT
Plug in the known values:
0.991 atm * V = 0.0912 mol * 0.0821 L·atm/(mol·K) * 297 K
Simplify the equation:
V = (0.0912 * 0.0821 * 297) / 0.991
Now solve for V:
V = 2.736 L
Therefore, the correct answer is:
V = 2.736 L
So, the volume of ammonia gas produced from 4.56 g of magnesium nitride at 24 degrees C and 753 mmHg is 2.736 L.
Let's check if your calculation is correct.
First, let's convert the grams of magnesium nitride to moles. You have correctly calculated that 4.56 g of magnesium nitride corresponds to 0.0912 mol of magnesium nitride using the molar mass of magnesium nitride.
Next, let's substitute the values into the ideal gas law equation: PV = nRT.
P = 0.991 atm
V = ?
n = 0.0912 mol
R = 0.0821 L·atm/(mol·K)
T = 297 K
Now we can solve for V. Rearranging the equation to solve for V gives us:
V = (nRT) / P
Substituting the values:
V = (0.0912 mol * 0.0821 L·atm/(mol·K) * 297 K) / 0.991 atm
Calculating this expression, we find:
V = 2.169 L
Therefore, the correct answer is V = 2.169 L. Your answer, V = 2.24 NH3, is not correct.