what will volume of ammonia gas when 100cm^3of nitrogen reacts with 150cm^3of hydrogen?which is the limiting reactant?
N2 + 3H2 ==> 2NH3
100 cc....150 cc........xxx
Determine the limiting reagent.
Using the coefficients like this:
100 cc N2 will produce 200 cc NH3 and
150 cc H2 will produce 150 x 2/3 = 100 cc NH3; therefore,H2 is the limiting reagent (LR) and N2 is the excess reagent (ER). 100 cc of NH3 gas will be produced.
To determine the volume of ammonia gas produced and identify the limiting reactant, we need to write and balance the chemical equation for the reaction between nitrogen and hydrogen to form ammonia.
The balanced chemical equation for the reaction is:
N2 + 3H2 -> 2NH3
According to the balanced equation, one mole of nitrogen reacts with three moles of hydrogen to produce two moles of ammonia.
Step 1: Convert the given volumes of nitrogen and hydrogen to moles using the ideal gas law equation:
n = V / Vm
Where:
n = moles
V = volume of gas in cm^3
Vm = molar volume of gas at standard temperature and pressure (STP), which is 22.4 L/mol (or 22,400 cm^3/mol)
Moles of nitrogen:
n(N2) = 100 cm^3 / 22,400 cm^3/mol
n(N2) ≈ 0.00446 mol
Moles of hydrogen:
n(H2) = 150 cm^3 / 22,400 cm^3/mol
n(H2) ≈ 0.00670 mol
Step 2: Determine the limiting reactant by comparing the mole ratios:
According to the balanced equation, the mole ratio of nitrogen to hydrogen is 1:3. This means that for ammonia production, one mole of nitrogen requires three moles of hydrogen.
Using the mole ratios, we can calculate the maximum moles of ammonia that can be produced from each reactant.
Maximum moles of ammonia from nitrogen:
0.00446 mol N2 × (2 mol NH3 / 1 mol N2) ≈ 0.00892 mol NH3
Maximum moles of ammonia from hydrogen:
0.00670 mol H2 × (2 mol NH3 / 3 mol H2) ≈ 0.00447 mol NH3
Based on the calculations, we can see that the moles of ammonia produced from nitrogen is higher than the moles produced from hydrogen. Therefore, nitrogen is the limiting reactant.
Step 3: Calculate the volume of ammonia gas produced:
From the balanced equation, we know that two moles of ammonia are produced for every one mole of nitrogen.
Volume of ammonia gas:
V(NH3) = n(NH3) × Vm
V(NH3) = 0.00892 mol × 22,400 cm^3/mol
V(NH3) ≈ 199.81 cm^3
Therefore, approximately 199.81 cm^3 of ammonia gas will be produced when 100 cm^3 of nitrogen reacts with 150 cm^3 of hydrogen.