you take the molar mass of ammonium sulfate (132.144) and multiply that by the 1.60 moles you are already given to get the answer, 211.4034 grams
the equation looks like this:
132.144*1.60=211.4304 g
This mass is 59,463 g.
80.043 g/mol
Mass of Nitrogen: 14g/mol Mass of Ammonium Nitrate (NH4NO3): 14 + 1x4 + 14 + 16x3 = 80g/mol ∴ % of Nitrogen in Ammonium Nitrate = 14/80 * 100 = 17.5%
152.10 g/mol
Ammonia is a compound made of Nitogen and Hydrogen with compound formula NH3 Ammonium is an ionized form of Ammonia which has the chemical symbol NH4.
This mass is 59,463 g.
14% FeThe formula for iron ammonium sulfate hexahydrate is (NH4)2Fe(SO4)2·6H2O, which can also be written as FeH20N2O14S2.To determine the percentage of Fe in the compound, divide the mass of Fe by the mass of the compound and multiply by 100. The molar mass of Fe is 56 g/mol. The molar mass of the compound is 392 g/mol. There is only one Fe atom in the formula, so the mass of Fe is 56 g/mol.%Fe = (56 g/mol)/(392 g/mol) x 100 = 14%
The mass in grams of 1,40 mol of anhydrous iron(III) sulfate is 559,832.
18.03846 g/mol
80.043 g/mol
211 is the mass in grams of 1.15 moles of strontium sulfate.
Mass of Nitrogen: 14g/mol Mass of Ammonium Nitrate (NH4NO3): 14 + 1x4 + 14 + 16x3 = 80g/mol ∴ % of Nitrogen in Ammonium Nitrate = 14/80 * 100 = 17.5%
Reactants: 30 g (NH4)2Cr2O7, 20 g MgSO4 30 g (NH4)2Cr2O7 / 252.07 g/mol = 0.119 mol 20 g MgSO4 / 120.41 g/mol = 0.166 mol Since they react in a 1:1 ratio to form magnesium dichromate and ammonium sulfate, ammonium dichromate is the limiting reagent (only 0.119 mol of MgSO4 is needed to react with all the (NH4)2Cr2O7).
152.10 g/mol
The molar mass for anhydrous barium sulfate (BaSO4) is 233.43 g/mol
Ammonia is a compound made of Nitogen and Hydrogen with compound formula NH3 Ammonium is an ionized form of Ammonia which has the chemical symbol NH4.
Nickel(II) sulfate