<u>Answer:</u> The value of
for
reaction is 
<u>Explanation:</u>
We are given:
Initial moles of nitrogen gas = 1.30 moles
Initial moles of hydrogen gas = 1.65 moles
Equilibrium moles of ammonia = 0.100 moles
Volume of the container = 1.00 L
For the given chemical equation:

<u>Initial:</u> 1.30 1.65
<u>At eqllm:</u> 1.30-x 1.65-3x 2x
Evaluating the value of 'x'

The expression of
for above equation follows:
![K_c=\frac{[NH_3]^2}{[N_2]\times [H_2]^3}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BNH_3%5D%5E2%7D%7B%5BN_2%5D%5Ctimes%20%5BH_2%5D%5E3%7D)
Equilibrium moles of nitrogen gas = 
Equilibrium moles of hydrogen gas = 
Putting values in above expression, we get:

Calculating the
for the given chemical equation:


Hence, the value of
for
reaction is 
Answer:
c is the answer then check it out
Answer:
C. The balloon with CH4 has the same moles of gas molecules as the balloon with H2
Explanation:
Based on combined gas law, gases under the same pressure, temperature and volume have the same number of moles. With this information we can say the rigth statement is:
<h3>C. The balloon with CH4 has the same moles of gas molecules as the balloon with H2</h3>
Answer:
Mass = 42.8g
Explanation:
4 NH 3 ( g ) + 5 O 2 ( g ) ⟶ 4 NO ( g ) + 6 H 2 O ( g )
Observe that every 4 mole of ammonia requires 5 moles of oxygen to obtain 4 moles of Nitrogen oxide and 6 moles of water.
Step 1: Determine the balanced chemical equation for the chemical reaction.
The balanced chemical equation is already given.
Step 2: Convert all given information into moles (through the use of molar mass as a conversion factor).
Ammonia = 63.4g × 1mol / 17.031 g = 3.7226mol
Oxygen = 63.4g × 1mol / 32g = 1.9813mol
Step 3: Calculate the mole ratio from the given information. Compare the calculated ratio to the actual ratio.
If all of the 1.9831 moles of oxygen were to be used up, there would need to be 1.9831 × 4 / 5 or 1.5865 moles of Ammonia. We have 3.72226 moles of ammonia - Far excess. Because there is an excess of Ammonia, the Oxygen amount is used to calculate the amount of the products in the reaction.
Step 4: Use the amount of limiting reactant to calculate the amount of H2O produced.
5 moles of O2 = 6 moles of H2O
1.9831 moles = x
x = (1.9831 * 6 ) / 5
x = 2.37972 moles
Mass of H2O = Molar mass * Molar mass
Mass = 2.7972 * 18
Mass = 42.8g