A powder contains FeSO4\u00b77H2O (molar mass=278.01 g\/mol), among other components. A 3.125 g sample of the powder was dissolved in HNO3 and heated to convert all iron to Fe3+. The addition of NH3 precipitated Fe2O3\u00b7xH2O, which was subsequently ignited to produce 0.469 g Fe2O3.<\/p>\n\n\n\n
What was the mass of FeSO4\u00b77H2O in the 3.125 g sample?<\/p>\n\n\n\n
mass of FeSO4\u00b77H2O:<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n Let’s work through this carefully:<\/p>\n\n\n\n First, given data<\/strong>:<\/p>\n\n\n\n Step 1: Relate Fe\u2082O\u2083 to Fe<\/strong><\/p>\n\n\n\n We know Fe\u2082O\u2083 contains iron (Fe). The molar mass of Fe\u2082O\u2083 is:<\/p>\n\n\n\n [ Now, find moles of Fe\u2082O\u2083:<\/p>\n\n\n\n [ Each mole of Fe\u2082O\u2083 contains 2 moles of Fe<\/strong>:<\/p>\n\n\n\n [ Step 2: Relate Fe to FeSO\u2084\u00b77H\u2082O<\/strong><\/p>\n\n\n\n In FeSO\u2084\u00b77H\u2082O, there is 1 mole of Fe per mole of FeSO\u2084\u00b77H\u2082O<\/strong>.<\/p>\n\n\n\n Thus:<\/p>\n\n\n\n [ Now, find the mass of FeSO\u2084\u00b77H\u2082O:<\/p>\n\n\n\n [ \u2705 Answer<\/strong>: This problem involves stoichiometry and the chemical transformation of iron from one compound to another. The powder initially contains FeSO\u2084\u00b77H\u2082O among other substances. When treated with nitric acid (HNO\u2083), all iron is oxidized to Fe\u00b3\u207a. Adding ammonia (NH\u2083) precipitates hydrated iron(III) oxide, Fe\u2082O\u2083\u00b7xH\u2082O. Heating removes the water, leaving behind pure Fe\u2082O\u2083.<\/p>\n\n\n\n We first use the mass of Fe\u2082O\u2083 obtained after ignition to find the number of moles of Fe\u2082O\u2083. Knowing the molar mass of Fe\u2082O\u2083 (159.69 g\/mol), the moles are calculated by dividing the mass (0.469 g) by the molar mass. Each mole of Fe\u2082O\u2083 contains exactly 2 moles of iron atoms because of the chemical formula (Fe\u2082O\u2083). Therefore, multiplying the moles of Fe\u2082O\u2083 by 2 gives the moles of Fe present in the original sample.<\/p>\n\n\n\n Since FeSO\u2084\u00b77H\u2082O contains one iron atom per formula unit, the number of moles of Fe corresponds directly to the number of moles of FeSO\u2084\u00b77H\u2082O originally present. Thus, we multiply the moles of FeSO\u2084\u00b77H\u2082O by its molar mass (278.01 g\/mol) to find the mass.<\/p>\n\n\n\n Finally, we find that 1.632 g<\/strong> of the 3.125 g sample was FeSO\u2084\u00b77H\u2082O. The rest of the sample consisted of other, unspecified components. This method accurately traces the quantity of the original iron-containing compound based on how much Fe\u2082O\u2083 was ultimately recovered, an essential technique in quantitative chemical analysis.<\/p>\n\n\n\n A powder contains FeSO4\u00b77H2O (molar mass=278.01 g\/mol), among other components. A 3.125 g sample of the powder was dissolved in HNO3 and heated to convert all iron to Fe3+. The addition of NH3 precipitated Fe2O3\u00b7xH2O, which was subsequently ignited to produce 0.469 g Fe2O3. What was the mass of FeSO4\u00b77H2O in the 3.125 g sample? 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\n\n\n\n\n
\n\n\n\n
M(\\text{Fe}_2\\text{O}_3) = 2(55.845) + 3(16.00) = 159.69\\ \\text{g\/mol}
]<\/p>\n\n\n\n
\\text{Moles of Fe}_2\\text{O}_3 = \\frac{0.469\\ \\text{g}}{159.69\\ \\text{g\/mol}} = 0.002936\\ \\text{mol}
]<\/p>\n\n\n\n
\\text{Moles of Fe} = 2 \\times 0.002936 = 0.005872\\ \\text{mol}
]<\/p>\n\n\n\n
\n\n\n\n
\\text{Moles of FeSO}_4\\cdot 7\\text{H}_2\\text{O} = \\text{Moles of Fe} = 0.005872\\ \\text{mol}
]<\/p>\n\n\n\n
\\text{Mass of FeSO}_4\\cdot 7\\text{H}_2\\text{O} = 0.005872\\ \\text{mol} \\times 278.01\\ \\text{g\/mol} = 1.632\\ \\text{g}
]<\/p>\n\n\n\n
\n\n\n\n
[
\\boxed{1.632\\ \\text{g}}
]<\/p>\n\n\n\n
\n\n\n\nExplanation (around 300 words):<\/h1>\n\n\n\n
\n","protected":false},"excerpt":{"rendered":"