a) Ammonia, an important source of fixed nitrogen that can be metabolized by plants, is produced using the Haber process in which nitrogen and hydrogen combine.
N2(g) + 3H2(g) → 2NH3(g)
How many grams of nitrogen are needed to produce 325 grams of ammonia?
b)A 0.00100 mol sample of Ca(OH)2 requires 25.00 mL of aqueous HCl for neutralization according to the reaction below. What is the concentration of the HCl?
Equation: Ca(OH)2(s) + 2HCl(aq) → CaCl2(aq) + H2O(l)
N 14 g/mol
H 1 g/mol
NH3 is therefore 17 g/mol, 2NH3 is 34
N2 is 28 g/mol
every 2 mol of NH3 requires 1 MOL OF N2
so
every 34 g of NH3 requires 28 g of N2
so finally
325 g NH3 (28 N2/34 NH3) = 268 g N2
every .001 mol Ca(OH)2 needs .002 mol of HCl
so
we have .002 mol in 25*10^-3 liters
.002/.025 = .08 mol/liter
asdf
a) To find out how many grams of nitrogen are needed to produce 325 grams of ammonia, we need to use the balanced equation provided.
The equation tells us that 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia.
First, we need to convert the given mass of ammonia (325 grams) to moles. To do this, we use the molar mass of ammonia (NH3), which is approximately 17 grams/mol.
Moles of NH3 = mass of NH3 / molar mass of NH3
Moles of NH3 = 325 grams / 17 grams/mol
Moles of NH3 ≈ 19.118 mol
Now, using the stoichiometric ratio from the balanced equation, we can determine the number of moles of nitrogen required to produce 19.118 moles of ammonia.
Moles of N2 = (Moles of NH3) / 2
Moles of N2 ≈ 19.118 mol / 2
Moles of N2 ≈ 9.559 mol
Finally, we can convert the moles of nitrogen to grams using the molar mass of nitrogen (N2), which is approximately 28 grams/mol.
Mass of N2 = moles of N2 × molar mass of N2
Mass of N2 ≈ 9.559 mol × 28 grams/mol
Mass of N2 ≈ 267.212 grams
Therefore, approximately 267.212 grams of nitrogen are needed to produce 325 grams of ammonia.
b) To find the concentration of HCl, we need to use the balanced equation provided and the given information.
The equation tells us that 1 mole of Ca(OH)2 reacts with 2 moles of HCl.
First, we need to convert the given amount of Ca(OH)2 (0.00100 mol) to moles of HCl. Since the stoichiometric ratio is 2:1, the moles of HCl will be the same as the moles of Ca(OH)2.
Moles of HCl = 0.00100 mol
Next, we need to find the volume of the HCl solution corresponding to the moles of HCl used. As given in the problem, it is 25.00 mL.
Now, we can calculate the concentration of the HCl solution.
Concentration (in mol/L) = Moles of solute / Volume of solution (in L)
Concentration = 0.00100 mol / 0.02500 L
Concentration = 0.040 mol/L
Therefore, the concentration of the HCl solution is 0.040 mol/L.