120 cm^3 of NH3 gas was prepared and collected at R.T.P. All the NH3 gas was reacted with H3PO4 according to the following reaction
3NH3(g) + H3PO4(aq)= (NH4)3 PO4(aq)
calculate:
1. the number of moles of NH3 that was collected
2. the mass of (NH4)3 PO4 obtained.
[N=14, O=16, P=31, H=1; molar gas volume at R.T.P. =24000 cm^3 ]
1. Use PV = nRT and solve for n = numer of mols.
2. mols NH3 from 1 = mols (NH4)3PO4
Then g (NH4)3PO4 = mols x molar mass
Post your work if you get stuck.
To calculate the number of moles of NH3 that was collected, we need to use the Ideal Gas Law equation, which is:
PV = nRT
Where:
P = pressure of the gas (R.T.P. = 1 atm)
V = volume of the gas (120 cm^3)
n = number of moles of the gas
R = ideal gas constant (0.0821 L⋅atm/(mol⋅K))
T = temperature in Kelvin (R.T.P. = 273 K)
First, let's convert the volume from cm^3 to liters (L):
120 cm^3 ÷ 1000 = 0.12 L
Now, we can rearrange the Ideal Gas Law equation to solve for the number of moles (n):
n = PV / RT
n = (1 atm)(0.12 L) / (0.0821 L⋅atm/(mol⋅K))(273 K)
n ≈ 0.005 mol
Therefore, the number of moles of NH3 collected is approximately 0.005 mol.
Now, to determine the mass of (NH4)3PO4 obtained, we need to use stoichiometry. From the balanced chemical equation, we can see that 3 moles of NH3 react with 1 mole of (NH4)3PO4.
Since we know the number of moles of NH3 collected (0.005 mol), we can use this information to calculate the moles of (NH4)3PO4 produced.
Moles of (NH4)3PO4 = (0.005 mol NH3) / (3 mol NH3/1 mol (NH4)3PO4)
Moles of (NH4)3PO4 ≈ 0.00167 mol
To determine the mass of (NH4)3PO4, we need to use its molar mass. Let's calculate it:
(NH4)3PO4:
3 nitrogen (N) atoms × 14 g/mol = 42 g/mol
12 hydrogen (H) atoms × 1 g/mol = 12 g/mol
1 phosphorus (P) atom × 31 g/mol = 31 g/mol
4 oxygen (O) atoms × 16 g/mol =64 g/mol
Total molar mass of (NH4)3PO4 = 42 g/mol + 12 g/mol + 31 g/mol + 64 g/mol = 149 g/mol
Now we can calculate the mass of (NH4)3PO4 obtained:
Mass of (NH4)3PO4 = Moles of (NH4)3PO4 × molar mass of (NH4)3PO4
Mass of (NH4)3PO4 = 0.00167 mol × 149 g/mol
Mass of (NH4)3PO4 ≈ 0.248 g
Therefore, the mass of (NH4)3PO4 obtained is approximately 0.248 g.