the important industrial process for producing ammonia (the Haber Process), the overall reaction is: N2(g) + 3H2(g) → 2NH3(g) + 100.4 kJ A yield of NH3 of approximately 98% can be obtained at 200°C and 1,000 atmospheres of pressure. How many grams of N2 must react to form 1.7 grams of ammonia, NH3
Is that 1.7 g NH3 at 98% or 1.7 g NH3 @ 98%? I will assume you want 1.7 g NH3 but with 98% yield.
N2 + 3H2 ==> 2NH3
mols NH3 = 1.7g/molar mass NH3 = 1.7/17 = 0.1 mol.
That will require how many mols N2? That's 0.1 mol NH3 x (1 mol N2/2 mol NH3) = 0.1 x 1/2 = 0.05 mol N2.
How many grams is that? That's 0.05 mol N2 x molar mass N2 = 0.05 x 28 = 1.4 g N2. That is assuming a 100% yield. If you have only 98% yield, you will need 1.4/0.98 = ? g N2. Check my work carefully.
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To determine the amount of N2 required to produce a given amount of ammonia, we need to use stoichiometry.
Step 1: Convert the given mass of ammonia (NH3) to moles.
Given mass of NH3 = 1.7 grams
Molar mass of NH3 = 17.03 g/mol (N: 14.01 g/mol, H: 1.01 g/mol x 3)
Using the formula:
moles of NH3 = given mass / molar mass
moles of NH3 = 1.7 g / 17.03 g/mol = 0.09988 mol (approximately 0.1 mol)
Step 2: Determine the mole ratio between NH3 and N2 from the balanced equation.
From the balanced equation, 1 mole of nitrogen (N2) reacts to form 2 moles of ammonia (NH3).
Using the ratio:
moles of N2 = moles of NH3 / 2
moles of N2 = 0.1 mol / 2 = 0.05 mol
Step 3: Convert moles of N2 to grams.
Molar mass of N2 = 28.01 g/mol
Using the formula:
mass of N2 = moles of N2 x molar mass
mass of N2 = 0.05 mol x 28.01 g/mol = 1.401 grams (approximately 1.4 g)
Therefore, approximately 1.4 grams of N2 must react to form 1.7 grams of ammonia (NH3).