Question

найти выход продукта аммиака если вступали в реакцию азот и 60г водорода образовалось 270г аммиака

Answer 1

n (NH3)= ?
m(H2) = 60 г
m (NH3) =270 г

Решение:
        60г     xг
N2 + 3H2 = 2NH3
      3 моль*    2 моль*
     *2 г/моль *17г/моль

n=m практ/m теор * 100%
M(NH3) = 14+3=17 г/моль
Складаємо пропорцію:
60/6=х/34
х=(60*34)/6=340 (г) - маса теоретична
Підставляємо у формулу:
n=(270/340)*100% = 79%

Відповідь: 79%

    

Answer 2

Finding the Yield of Ammonia

To find the yield of ammonia (NH3) when nitrogen (N2) and 60g of hydrogen (H2) react to form 270g of ammonia, we can use the concept of stoichiometry. Stoichiometry is a branch of chemistry that deals with the quantitative relationships between reactants and products in chemical reactions.

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

N2 + 3H2 → 2NH3

From the balanced equation, we can see that 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia. We can use this information to calculate the theoretical yield of ammonia.

Calculating the Theoretical Yield

To calculate the theoretical yield of ammonia, we need to determine the number of moles of nitrogen and hydrogen in the given amounts.

1. Calculate the number of moles of nitrogen (N2): - The molar mass of nitrogen (N2) is 28 g/mol. - Divide the given mass of nitrogen (N2) by its molar mass to find the number of moles: - Moles of N2 = Mass of N2 / Molar mass of N2 - Moles of N2 = 0 g (since nitrogen is not given in the question)

2. Calculate the number of moles of hydrogen (H2): - The molar mass of hydrogen (H2) is 2 g/mol. - Divide the given mass of hydrogen (H2) by its molar mass to find the number of moles: - Moles of H2 = Mass of H2 / Molar mass of H2 - Moles of H2 = 60 g / 2 g/mol - Moles of H2 = 30 mol

3. Determine the limiting reactant: - The limiting reactant is the reactant that is completely consumed in the reaction and determines the maximum amount of product that can be formed. - To determine the limiting reactant, we compare the moles of nitrogen and hydrogen. - Since the balanced equation shows that 1 mole of nitrogen reacts with 3 moles of hydrogen, we can see that the ratio of nitrogen to hydrogen is 1:3. - Since we have 30 moles of hydrogen, we would need 10 moles of nitrogen to react completely (30 moles of H2 / 3 moles of H2 per 1 mole of N2). - However, we don't have any nitrogen given in the question, so nitrogen is the limiting reactant.

4. Calculate the theoretical yield of ammonia: - Since nitrogen is the limiting reactant, the amount of ammonia formed will be determined by the moles of nitrogen. - From the balanced equation, we know that 1 mole of nitrogen reacts to form 2 moles of ammonia. - Therefore, the theoretical yield of ammonia can be calculated as follows: - Moles of NH3 = Moles of N2 × (2 moles of NH3 / 1 mole of N2) - Moles of NH3 = 10 mol × (2 mol NH3 / 1 mol N2) - Moles of NH3 = 20 mol

5. Convert the moles of ammonia to grams: - The molar mass of ammonia (NH3) is 17 g/mol. - Multiply the moles of ammonia by its molar mass to find the mass of ammonia: - Mass of NH3 = Moles of NH3 × Molar mass of NH3 - Mass of NH3 = 20 mol × 17 g/mol - Mass of NH3 = 340 g

Therefore, the theoretical yield of ammonia when nitrogen and 60g of hydrogen react is 340g.

Conclusion

In conclusion, when nitrogen and 60g of hydrogen react, the theoretical yield of ammonia is 340g