Answer:
Therewill be produced 170.6 grams NH3, there will remain 25 moles of N2, this is 700 grams
Explanation:
<u>Step 1:</u> 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
<u>Step 2:</u> The balanced equation
N2 + 3H2 → 2NH3
<u>Step 3:</u> 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.
<u>Step 4:</u> 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
<u>Step 5</u>: 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)
<u>Step 6: </u>Calculate actual mass
50% yield = actual mass / theoretical mass
actual mass = 0.5 * 340.6
actual mass = 170.3 grams
<u>Step 7:</u> 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
