what volume of nitrogen at 215 degrees celsius and 715mm Hg would be required to react with 0.100 mol of hydrogen to produce ammonia? N2(g) + 3H2(g)-->2NH3(g)? Any help is greatly appreciated.

0.1 mole H2 will require how many moles N2 at STP? That will be 0.1 mol H2 x (1 mole N2/3 moles H2) = 0.1 mol x (1/3) = ?moles N2. What volume is that. ?mol x 22.4L/mol = ?L H2. Now convert that to the conditions listed with PV = nRT.

Well, it seems like you're looking for the volume of nitrogen needed to react with hydrogen to produce ammonia. Let's calculate it, but first, I want to assure you I won't make any exploding jokes. Nitrogen is quite stable, you know.

Given the balanced equation is N2(g) + 3H2(g) → 2NH3(g), we know that one mole of nitrogen reacts with three moles of hydrogen to produce two moles of ammonia. Right? Right.

Now, we have 0.100 mol of hydrogen. Since the ratio with nitrogen is 3:1, we'll need one-third of that amount of nitrogen, which is 0.033 mol of nitrogen.

So, all that's left is to calculate the volume. For this, we'll be using the Ideal Gas Law equation: PV = nRT. But hang on, I'll do the math for you.

Given:
T = 215°C = 215 + 273 = 488 K
P = 715 mm Hg

Now, let's convert the pressure to atmospheres (atm). 1 atm = 760 mm Hg, so 715 mm Hg = 715/760 = 0.94 atm.

Using the Ideal Gas Law equation:

(V)(0.94 atm) = (0.033 mol)(0.0821 L·atm/mol·K)(488 K)

Solving for V, we get:

V ≈ (0.033)(0.0821)(488) / (0.94)

Calculating this, we find:

V ≈ 1.002 L

So, approximately 1.002 liters of nitrogen will be required at 215 degrees Celsius and 715 mm Hg to react with 0.100 mol of hydrogen to produce ammonia. Voilà!

Remember, this is just an approximation, so don't hold me accountable for any slight variances.

To find the volume of nitrogen, we can use the ideal gas law equation: PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

First, we need to convert the given temperature from degrees Celsius to Kelvin. To convert Celsius to Kelvin, we use the equation K = °C + 273.15.

Given:
Number of moles of hydrogen (H2) = 0.100 mol
Temperature (T) = 215 degrees Celsius = 215 + 273.15 = 488.15 K
Pressure (P) = 715 mmHg

We also need to know the molar ratio of nitrogen to hydrogen in the balanced chemical equation. From the equation: N2(g) + 3H2(g) --> 2NH3(g), we can see that the ratio is 1 mol of nitrogen reacts with 3 mol of hydrogen.

Now we can proceed to calculate the volume of nitrogen gas.

Step 1: Convert pressure from mmHg to atm.
1 atm = 760 mmHg
Pressure (P) = 715 mmHg ÷ 760 mmHg/atm ≈ 0.9408 atm

Step 2: Use the ideal gas law equation to find the volume of nitrogen gas.
PV = nRT
V = (nRT) / P
V = (0.100 mol * 0.0821 atm L/mol K * 488.15 K) / 0.9408 atm
V ≈ 4.244 L

Therefore, approximately 4.244 liters of nitrogen gas would be required to react with 0.100 mol of hydrogen to produce ammonia.

To solve this problem, we need to use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

First, let's calculate the number of moles of nitrogen required to react with 0.100 mol of hydrogen. According to the balanced chemical equation, 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia. Therefore, if we have 0.100 mol of hydrogen, we need (0.100 mol * 1 mol of N2 / 3 mol of H2) = 0.0333 mol of nitrogen.

Now, let's convert the temperature from Celsius to Kelvin. The Kelvin temperature is obtained by adding 273.15 to the Celsius temperature. So, 215°C + 273.15 = 488.15 K.

Next, we can substitute the values into the ideal gas law equation to solve for the volume of nitrogen (V). The gas constant (R) is typically expressed as 0.0821 L·atm/(K·mol).

PV = nRT
(V)(715 mm Hg) = (0.0333 mol)(0.0821 L·atm/(K·mol))(488.15 K)

Now, let's convert the pressure from mm Hg to atm by dividing by 760 mm Hg/atm.

(V)(715 mm Hg / 760 mm Hg/atm) = (0.0333 mol)(0.0821 L·atm/(K·mol))(488.15 K)

Simplifying the equation,

V = [(0.0333 mol)(0.0821 L·atm/(K·mol))(488.15 K)] / (715 mm Hg / 760 mm Hg/atm)

Now we can solve for V:

V ≈ 0.031 L

Therefore, approximately 0.031 L of nitrogen would be required to react with 0.100 mol of hydrogen at 215°C and 715 mm Hg to produce ammonia.