1 kg of N2 is mixed with 3.5 m^3 of H2 at 300k,101.3kpa and sent to ammonia converter , product analysed 13.7℅ NH3,70.32℅H2,15.98℅ N2 find ℅ conversion, ℅excess
To find the percent conversion and percent excess in this question, we need to compare the amount of reactants before and after the reaction. Here's how to proceed:
1. Determine the initial moles of N2 and H2:
- We are given that the mass of N2 is 1 kg.
- The molar mass of N2 is approximately 28 g/mol.
- Therefore, the initial moles of N2 can be calculated as follows:
Initial moles of N2 = mass of N2 / molar mass of N2
= 1000 g / 28 g/mol
= 35.71 mol
- We are given the volume of H2 as 3.5 m^3.
- The ideal gas law can be used to calculate the number of moles of H2:
PV = nRT, where P is pressure, V is volume, n is moles, R is the ideal gas constant, and T is temperature in Kelvin.
Rearranging the equation:
n = PV / RT
n = (101.3 kPa) * (3.5 m^3) / [(8.314 J/(mol·K)) * (300 K)]
n ≈ 1.62 x 10^4 mol
2. Determine the limiting reactant:
- The balanced chemical equation for the reaction is:
N2 + 3H2 -> 2NH3
- According to the stoichiometry, 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.
- Comparing the initial moles of N2 (35.71 mol) and H2 (1.62 x 10^4 mol), we can see that N2 is the limiting reactant because it is present in fewer moles.
3. Determine the moles of NH3 produced:
- Since 2 moles of NH3 are produced for every 1 mole of N2, the moles of NH3 can be calculated as follows:
Moles of NH3 = 2 * initial moles of N2
= 2 * 35.71 mol
= 71.42 mol
4. Determine the total moles of products:
- The total moles of products are the sum of the moles of NH3 and H2 present after the reaction:
Total moles of products = moles of NH3 + moles of H2
= 71.42 mol + (0.7032 * 1.62 x 10^4 mol) (based on the percentage composition of H2 after reaction)
5. Calculate the percent conversion:
- The percent conversion can be calculated as the ratio of moles of NH3 produced to the total moles of products, multiplied by 100:
Percent conversion = (moles NH3 / total moles of products) * 100
= (71.42 mol / (71.42 mol + (0.7032 * 1.62 x 10^4 mol)) * 100
6. Calculate the percent excess:
- The percent excess can be calculated as the ratio of the excess moles of H2 to the total moles of products, multiplied by 100:
Percent excess = (excess moles of H2 / total moles of products) * 100
= ((0.7032 * 1.62 x 10^4 mol) / (71.42 mol + (0.7032 * 1.62 x 10^4 mol))) * 100
By following these steps, you should be able to find the percent conversion and percent excess for the given reaction.