Answer:
a) N2 is the limiting reactant
b) <u>1215.9 grams NH3</u>
<u>c) 2283.6 grams H2</u>
Explanation:
Step 1: Data given
Mass of N2 = 1000 grams
Molar mass N2 = 28 g/mol
Mass of H2 = 2500 grams
Molar mass H2 = 2.02 g/mol
Step 2: The balanced equation
N2(g) +3H2(g) → 2NH3(g)
Step 3: Calculate moles N2
Moles N2 = mass N2 / molar mass N2
Moles N2 = 1000 grams / 28.0 g/mol
Moles N2 = 35.7 moles
Step 4: Calculate moles H2
Moles H2 = 2500 grams / 2.02 g/mol
Moles H2 = 1237.6 moles
Step 5: Calculate limiting reactant
For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3
<u>N2 is the limiting reactant</u>. There will be consumed 35.7 moles.
H2 is in excess. There will react 3*35.7 = 107.1 moles
There will remain 1237.6 - 107.1= 1130.5 moles H2
This is 1130.5 * 2.2 = <u>2283.6 grams H2</u>
Step 6: Calculate moles NH3
For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3
For 35.7 moles N2 we'll have 2*35.7 = 71.4 moles NH3
Step 7: Calculate mass NH3
Mass NH3 = moles NH3 * molar mass NH3
Mass NH3 = 71.4 moles * 17.03 g/mol
Mass NH3 = <u>1215.9 grams NH3</u>
