Determine the mass of ammonium sulphate that would be produced from 68 grams of ammonium
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Chemistry
To determine the mass of ammonium sulfate produced from 68 grams of ammonium, we need to consider the stoichiometry of the chemical reaction between ammonium and ammonium sulfate.
The balanced chemical equation for the reaction involving ammonium sulfate is as follows:
2 NH4+ + SO4²- → (NH4)2SO4
From the equation, we can see that two ammonium ions (NH4+) combine with one sulfate ion (SO4²-) to form one molecule of ammonium sulfate ((NH4)2SO4).
To find the molar mass of ammonium sulfate, we need to sum up the atomic masses of all the atoms in the formula unit. The molar mass of ammonium sulfate (NH4)2SO4 can be calculated as follows:
(2 × molar mass of NH4+) + molar mass of SO4²-
The molar mass of nitrogen (N) is 14.01 g/mol, hydrogen (H) is 1.01 g/mol, sulfur (S) is 32.06 g/mol, and oxygen (O) is 16.00 g/mol.
So, the molar mass of ammonium sulfate is:
(2 × 14.01 g/mol) + 32.06 g/mol + (4 × 1.01 g/mol) + (4 × 16.00 g/mol)
= 132.14 g/mol
Now we can use this information to determine the mass of ammonium sulfate produced.
First, calculate the number of moles of ammonium (NH4+):
moles of NH4+ = mass of ammonium / molar mass of NH4+
moles of NH4+ = 68 g / 14.01 g/mol
moles of NH4+ ≈ 4.854 mol
Since the stoichiometry of the reaction is 2:1, we need to divide the number of moles of ammonium by 2 to find the number of moles of ammonium sulfate produced:
moles of (NH4)2SO4 = moles of NH4+ / 2
moles of (NH4)2SO4 = 4.854 mol / 2
moles of (NH4)2SO4 ≈ 2.427 mol
Finally, calculate the mass of ammonium sulfate produced:
mass of (NH4)2SO4 = moles of (NH4)2SO4 × molar mass of (NH4)2SO4
mass of (NH4)2SO4 = 2.427 mol × 132.14 g/mol
mass of (NH4)2SO4 ≈ 320.2 g
Therefore, approximately 320.2 grams of ammonium sulfate would be produced from 68 grams of ammonium.