Chemistry - Stoichiometry Question with Solution | TestHub

ChemistryStoichiometryProblems Based on Mixtures, Percentage PurityEasy2 minQB
ChemistryEasysingle choice
Passage / Comprehension
{"root":{"type":"root","format":"","indent":0,"version":1,"children":[{"type":"paragraph","format":"","indent":0,"version":1,"children":[{"mode":"normal","text":"Chromium exists as ","type":"text","style":"","detail":0,"format":0,"version":1},{"type":"equation","format":"latex","inline":true,"version":1,"equation":"\\mathrm{FeCr_{2}O_{4}}"},{"mode":"normal","text":" in the nature and it contains ","type":"text","style":"","detail":0,"format":0,"version":1},{"type":"equation","format":"latex","inline":true,"version":1,"equation":"\\mathrm{Fe_{0.95}O_{1.00}}"},{"mode":"normal","text":" as an impurity. To obtain pure chromium from ","type":"text","style":"","detail":0,"format":0,"version":1},{"type":"equation","format":"latex","inline":true,"version":1,"equation":"\\mathrm{FeCr_{2}O_{4}}"},{"mode":"normal","text":", the ore is fused with ","type":"text","style":"","detail":0,"format":0,"version":1},{"type":"equation","format":"latex","inline":true,"version":1,"equation":"\\mathrm{KOH}"},{"mode":"normal","text":" and oxygen is passed through the mixture when ","type":"text","style":"","detail":0,"format":0,"version":1},{"type":"equation","format":"latex","inline":true,"version":1,"equation":"\\mathrm{K_{2}CrO_{4}}"},{"mode":"normal","text":" and ","type":"text","style":"","detail":0,"format":0,"version":1},{"type":"equation","format":"latex","inline":true,"version":1,"equation":"\\mathrm{Fe_{2}O_{3}}"},{"mode":"normal","text":" are produced. A quantity of 2 g of ore required 336 ml of ","type":"text","style":"","detail":0,"format":0,"version":1},{"type":"equation","format":"latex","inline":true,"version":1,"equation":"\\mathrm{O_{2}}"},{"mode":"normal","text":" at 0°C and 1 atm for complete oxidation of ore. ","type":"text","style":"","detail":0,"format":0,"version":1},{"type":"equation","format":"latex","inline":true,"version":1,"equation":"\\mathrm{K_{2}CrO_{4}}"},{"mode":"normal","text":" is then precipitated as ","type":"text","style":"","detail":0,"format":0,"version":1},{"type":"equation","format":"latex","inline":true,"version":1,"equation":"\\mathrm{BaCrO_{4}}"},{"mode":"normal","text":" after addition of Barium salt. To the remaining solution, 12 ml of 1 M - ","type":"text","style":"","detail":0,"format":0,"version":1},{"type":"equation","format":"latex","inline":true,"version":1,"equation":"\\mathrm{K_{4}Fe(CN)_{6}}"},{"mode":"normal","text":" is added when ","type":"text","style":"","detail":0,"format":0,"version":1},{"type":"equation","format":"latex","inline":true,"version":1,"equation":"\\mathrm{Fe^{3+}}"},{"mode":"normal","text":" ions reacts with it to form ","type":"text","style":"","detail":0,"format":0,"version":1},{"type":"equation","format":"latex","inline":true,"version":1,"equation":"\\mathrm{KFe[Fe(CN)_{6}}"},{"mode":"normal","text":"], after called 'Prussian Blue'. To determine excess of ","type":"text","style":"","detail":0,"format":0,"version":1},{"type":"equation","format":"latex","inline":true,"version":1,"equation":"\\mathrm{K_{4}Fe(CN)_{6}}"},{"mode":"normal","text":" in solution, 9.25 ml of 0.4 N - ","type":"text","style":"","detail":0,"format":0,"version":1},{"type":"equation","format":"latex","inline":true,"version":1,"equation":"\\mathrm{Fe^{2+}}"},{"mode":"normal","text":" is added when all the ","type":"text","style":"","detail":0,"format":0,"version":1},{"type":"equation","format":"latex","inline":true,"version":1,"equation":"\\mathrm{K_{4}Fe(CN)_{6}}"},{"mode":"normal","text":" is precipitated as ","type":"text","style":"","detail":0,"format":0,"version":1},{"type":"equation","format":"latex","inline":true,"version":1,"equation":"\\mathrm{K_{2}Fe[Fe(CN)_{6}}"},{"mode":"normal","text":"]. (Fe = 56)","type":"text","style":"","detail":0,"format":0,"version":1}],"direction":"ltr","textFormat":0}],"direction":"ltr"}}

What is the percentage of in the ore?

Options:

Answer:
A
Solution:

Let g = (M = 224, 7 per formula unit) and g = (M = 69.2, 0.85 per formula unit). = 336/22400 = 15 mmol = 60 meq: 7x/224 + 0.85y/69.2 = 0.06, with + = 2. Solving: 0.0625 - 0.03125y + 0.01228y = 0.06, so = 0.0025/0.01897 = 0.1318 g. Percentage = 0.1318/2 × 100 = 6.59%, matching (a).

Stream:JEESubject:ChemistryTopic:StoichiometrySubtopic:Problems Based on Mixtures, Percentage Purity
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