Determine the mass of ammonium sulphate that would be produced from 68g of ammonia?
first, what is the reaction equation?
change 68g NH3 to moles
then check the equation to find the mole ratio of product to NH3
convert that back to grams
help me to answering this question
68g NH3 to moles?
NH3 has mol wt of 14+3=17
so, 68g = 4 moles
Now, what is the balanced equation?
That will tell you the relative amounts (in moles) of the reactants and products.
To determine the mass of ammonium sulfate produced from 68g of ammonia, we need to first write and balance the chemical equation for the reaction between ammonia (NH3) and sulfuric acid (H2SO4). The balanced equation is as follows:
2NH3 + H2SO4 -> (NH4)2SO4
From the balanced equation, we can see that 2 moles of ammonia react with 1 mole of sulfuric acid to produce 1 mole of ammonium sulfate.
Now, we need to calculate the molar mass of ammonia (NH3) and ammonium sulfate ((NH4)2SO4). The molar mass of ammonia is:
(N: 14.01 g/mol) + (H: 1.01 g/mol x 3) = 17.03 g/mol
The molar mass of ammonium sulfate is:
(N: 14.01 g/mol x 2) + (H: 1.01 g/mol x 8) + (S: 32.07 g/mol) + (O: 16.00 g/mol x 4) = 132.14 g/mol
Since we know the molar ratio from the balanced equation (2 moles of ammonia to 1 mole of ammonium sulfate), we can set up a proportion to find the mass of ammonium sulfate formed.
(68 g of ammonia) * (1 mol of (NH4)2SO4 / 2 mol of NH3) * (132.14 g/mol of (NH4)2SO4) = 224.89 g
Therefore, the mass of ammonium sulfate produced from 68g of ammonia is approximately 224.89 grams.