In Haber's process, 30 moles of hydrogen and 30 moles of nitrogen react to make ammonia. If the yield of the product is 50%, what is the mass of nitrogen remaining after the reaction?

25 moles

Explanation:

1 mole of nitrogen reacts with 3 moles of hydrogen to give 2 moles of ammonia. 30 moles of hydrogen will react with 10 moles of nitrogen to give 20 moles of ammonia. As the actual yield is 50%, ammonia formed is 10 moles, the amount of nitrogen reacted is 5 moles, and the amount of hydrogen reacted is 15 moles. The mass of the remaining hydrogen is 15 moles and of the remaining nitrogen is 25 moles.

Therewill be produced 170.6 grams NH3, there will remain 25 moles of N2, this is 700 grams

Explanation:

Step 1: Data given

Number of moles hydrogen = 30 moles

Number of moles nitrogen = 30 moles

Yield = 50 %

Molar mass of N2 = 28 g/mol

Molar mass of H2 = 2.02 g/mol

Molar mass of NH3 = 17.03 g/mol

Step 2: The balanced equation

N2 + 3H2 → 2NH3

Step 3: Calculate limiting reactant

For 1 mol of N2, we need 3 moles of H2 to produce 2 moles of NH3

Hydrogen is the limiting reactant.

The 30 moles will be completely be consumed.

N2 is in excess. There will react 30/3 = 10 moles

There will remain 30 - 10 = 20 moles (this in the case of a 100% yield)

In a 50 % yield, there will remain 20 + 0,5*10 = 25 moles. there will react 5 moles.

Step 4: Calculate moles of NH3

There will be produced, 30 / (3/2) = 20 moles of NH3 (In case of 100% yield)

For a 50% yield there will be produced, 10 moles of NH3

Step 5: Calculate the mass of NH3

Mass of NH3 = mol NH3 * Molar mass NH3

Mass of NH3 = 20 moles * 17.03

Mass of NH3 = 340.6 grams = theoretical yield (100% yield)

Step 6: Calculate actual mass

50% yield = actual mass / theoretical mass

actual mass = 0.5 * 340.6

actual mass = 170.3 grams

Step 7: The mass of nitrogen remaining

There remain 20 moles of nitrogen + 50% of 10 moles = 25 moles remain

Mass of nitrogen = 25 moles * 28 g/mol

Mass of nitrogen = 700 grams