A sample is a mixture of Mohr's salt and (NH4)2SO4 A 0 5 gram sample of the mixture gave - Chemistry - Structure of Atom

Question

A sample is a mixture of Mohr's salt and (NH4)2SO4 A 0 5 gram
sample of the mixture gave a precipitate of 0 75 grams of Barium
sulphate Calculate the percentage composition of the mixture What
weight of Fe2O3 be obtained if 0 2 grams of the mixture were
ignited in air?

Prerna Ahuja · Accepted Answer

Dear Student,

When the mixture of mohr's salt and ammonium sulphate reacts with
barium chloride, only ammonium sulphate reacts to form barium
sulphate The reaction being:

NH4SO4 + BaCl2 → BaSO4 + 2NH4ClFrom the equation, it can be seen that;For 1 mole of BaSO4, 1 mole of NH4SO4 is required
i e  For 233 38g of BaSO4, 132
14g of NH4SO4 is required
According to the question, 0
75 gm of BaSO4 was formedSo, For  0
75g of BaSO4, 132 14233 38× 0
75g of NH4SO4 is required=0
42g of NH4SO4 is requiredsince the mixture was 0
5 g,amount of mohr salt = 0
5 - 0 42 = 0
08gpercentage composition of mohr's salt = 0
080
5×100 = 16%percentage composition of  NH4SO4 = 100 - 16 = 84%

A sample is a mixture of Mohr's salt and (NH4)2SO4. A 0.5 gram sample of the mixture gave a precipitate of 0.75 grams of Barium sulphate. Calculate the percentage composition of the mixture. What weight of Fe2O3 be obtained if 0.2 grams of the mixture were ignited in air?