10 L of nitrogen gas and 10 L of hydrogen gas are introduced into an evacuated flask of 10 L capacity . it is then heated to 700 K and 3 atm pressure. what volume of ammonia at (NTP) is produced?

Question

Answer

Calculation of the volume of ammonia produced

Step 1: Write the balanced chemical equation

The balanced chemical equation for the reaction between nitrogen and hydrogen to form ammonia is:

N2(g) + 3H2(g) → 2NH3(g)

Step 2: Calculate the limiting reactant

To determine the limiting reactant, we must calculate the number of moles of each gas present in the flask.

Number of moles of nitrogen gas (N2):

PV = nRT

n = PV/RT = (3 atm x 10 L)/(0.0821 L atm/mol K x 700 K) = 0.427 mol

Number of moles of hydrogen gas (H2):

PV = nRT

n = PV/RT = (3 atm x 10 L)/(0.0821 L atm/mol K x 700 K) = 0.427 mol

Since the molar ratio of N2 to H2 in the balanced equation is 1:3, the limiting reactant is nitrogen gas because it will be completely consumed before all of the hydrogen gas is used up.

Step 3: Calculate the number of moles of ammonia produced

Using the balanced chemical equation, we know that 1 mole of nitrogen gas reacts with 3 moles of hydrogen gas to form 2 moles of ammonia gas. Therefore, the number of moles of ammonia produced is:

0.427 mol N2 x (2 mol NH3/1 mol N2) = 0.854 mol NH3

Step 4: Calculate the volume of ammonia produced at NTP

At NTP (normal temperature and pressure), which is defined as 0°C and 1 atm, the volume of 1 mole of any gas is 22.4 L. Therefore, the volume of ammonia produced is:

0.854 mol x 22.4 L/mol = 19.1 L

Therefore, the volume of ammonia produced at NTP is 19.1 L.

Answer

N2 + 3H2 --> 2NH3
10L 10L (?)

From the above balanced chemical reaction, H2 is the limiting reagent.

3 moles of H2 will give 2 moles of NH3.

Now, we have
T= 700K and P=3atm

appling PV=NRT,

700*V=2*0.082*700
V=57.4 L

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