Chemistry - Stoichiometry Question with Solution | TestHub
ChemistryStoichiometryProblems Based on Mixtures, Percentage PurityMedium2 minQB
ChemistryMediumsingle choice
Passage / Comprehension
{"root":{"type":"root","format":"","indent":0,"version":1,"children":[{"type":"paragraph","format":"","indent":0,"version":1,"children":[{"mode":"normal","text":"Chile saltpeter, a source of ","type":"text","style":"","detail":0,"format":0,"version":1},{"type":"equation","format":"latex","inline":true,"version":1,"equation":"\\mathrm{NaNO_{3}}"},{"mode":"normal","text":", also contains ","type":"text","style":"","detail":0,"format":0,"version":1},{"type":"equation","format":"latex","inline":true,"version":1,"equation":"\\mathrm{NaIO_{3}}"},{"mode":"normal","text":". The ","type":"text","style":"","detail":0,"format":0,"version":1},{"type":"equation","format":"latex","inline":true,"version":1,"equation":"\\mathrm{NaIO_{3}}"},{"mode":"normal","text":" can be used as a source of iodine, produced in the following reactions: ","type":"text","style":"","detail":0,"format":0,"version":1},{"type":"equation","format":"latex","inline":true,"version":1,"equation":"\\mathrm{IO_{3}^{-}}"},{"mode":"normal","text":" + ","type":"text","style":"","detail":0,"format":0,"version":1},{"type":"equation","format":"latex","inline":true,"version":1,"equation":"\\mathrm{3HSO_{3}^{-}}"},{"mode":"normal","text":" ","type":"text","style":"","detail":0,"format":0,"version":1},{"type":"equation","format":"latex","inline":true,"version":1,"equation":"\\rightarrow "},{"mode":"normal","text":" ","type":"text","style":"","detail":0,"format":0,"version":1},{"type":"equation","format":"latex","inline":true,"version":1,"equation":"\\mathrm{I^{-}}"},{"mode":"normal","text":" + ","type":"text","style":"","detail":0,"format":0,"version":1},{"type":"equation","format":"latex","inline":true,"version":1,"equation":"\\mathrm{3H^{+}}"},{"mode":"normal","text":" + ","type":"text","style":"","detail":0,"format":0,"version":1},{"type":"equation","format":"latex","inline":true,"version":1,"equation":"\\mathrm{3SO_{4}^{2-}}"},{"mode":"normal","text":" and ","type":"text","style":"","detail":0,"format":0,"version":1},{"type":"equation","format":"latex","inline":true,"version":1,"equation":"\\mathrm{5I^{-}}"},{"mode":"normal","text":" + ","type":"text","style":"","detail":0,"format":0,"version":1},{"type":"equation","format":"latex","inline":true,"version":1,"equation":"\\mathrm{IO_{3}^{-}}"},{"mode":"normal","text":" + ","type":"text","style":"","detail":0,"format":0,"version":1},{"type":"equation","format":"latex","inline":true,"version":1,"equation":"\\mathrm{6H^{+}}"},{"mode":"normal","text":" ","type":"text","style":"","detail":0,"format":0,"version":1},{"type":"equation","format":"latex","inline":true,"version":1,"equation":"\\rightarrow "},{"mode":"normal","text":" ","type":"text","style":"","detail":0,"format":0,"version":1},{"type":"equation","format":"latex","inline":true,"version":1,"equation":"\\mathrm{3I_{2}}"},{"mode":"normal","text":" + ","type":"text","style":"","detail":0,"format":0,"version":1},{"type":"equation","format":"latex","inline":true,"version":1,"equation":"\\mathrm{3H_{2}O}"},{"mode":"normal","text":". Five hundred millilitres of Chile saltpeter solution containing 4.95 g ","type":"text","style":"","detail":0,"format":0,"version":1},{"type":"equation","format":"latex","inline":true,"version":1,"equation":"\\mathrm{NaIO_{3}}"},{"mode":"normal","text":" is treated with stoichiometric quantity of ","type":"text","style":"","detail":0,"format":0,"version":1},{"type":"equation","format":"latex","inline":true,"version":1,"equation":"\\mathrm{NaHSO_{3}}"},{"mode":"normal","text":". Now an additional amount of same solution is added to the reaction mixture to bring about the second reaction. (I = 127)","type":"text","style":"","detail":0,"format":0,"version":1}],"direction":"ltr","textFormat":0}],"direction":"ltr"}}
How many grams of can be produced per litre of Chile saltpeter?
Options:
Answer:
B
Solution:
Per litre the solution holds 0.05 mol . Step I gives 0.05 mol , and step II converts it (with the extra 0.01 mol ) to : 0.05 × 3/5 = 0.03 mol . Mass = 0.03 × 254 = 7.62 g, matching (b).
Stream:JEESubject:ChemistryTopic:StoichiometrySubtopic:Problems Based on Mixtures, Percentage Purity
⏱ 2mℹ️ Source: QB
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