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Chapter 4: Problem 96

Solid iron(III) hydroxide is added to \(625 \mathrm{~mL}\) of \(0.280 \mathrm{M} \mathrm{HCl}\). The resulting solution is acidic and titrated with \(238.2 \mathrm{~mL}\) of \(0.113 \mathrm{M} \mathrm{NaOH}\). What mass of iron(III) hydroxide was added to the HCl?

Short Answer

Expert verified
Answer: The mass of iron(III) hydroxide added to the HCl solution is approximately 0.958 g.

Step by step solution

01

Determine the moles of HCl

We are given the initial concentration and volume of HCl. We can use the formula \(moles = (concentration)(volume)\) to find the total moles of HCl: \(moles_{HCl} = (0.280~M)(0.625~L) = 0.175~mol\)
02

Determine the moles of NaOH used in titration

We are given the volume and concentration of NaOH used for titration. We will use the same formula as used for the HCl: \(moles_{NaOH} = (0.113~M)(0.2382~L) = 0.02690~mol\)
03

Determine the remaining moles of HCl after iron(III) hydroxide addition

Knowing the moles of NaOH used in titration, we can find out how many moles of HCl remained unreacted with iron(III) hydroxide. To do that, we can use the stoichiometry of the reaction between HCl and NaOH, which is a 1:1 reaction: \(0.175~mol - 0.02690~mol = 0.1481~mol\) Hence, there were 0.1481 moles of HCl remaining unreacted after the addition of iron(III) hydroxide.
04

Determine moles of iron(III) hydroxide contributed to reaction

Now, we can find the moles of iron(III) hydroxide that reacted with HCl before titration. This can be done as we know that the stoichiometry of the reaction between HCl and iron(III) hydroxide: \(Fe(OH)_3 + 3HCl \rightarrow FeCl_3 + 3H_2O\). For every mole of \(Fe(OH)_3\), three moles of HCl are reacted: \(moles_{Fe(OH)_3} = (0.175~mol - 0.1481~mol) / 3 = 0.00897~mol\)
05

Calculate the mass of iron(III) hydroxide added

Finally, we can find the mass of iron(III) hydroxide added to HCl. We know the molar mass of \(Fe(OH)_3\), which can be calculated as follows: \(Fe(OH)_3: 1(55.845) + 3(16 + 1.008) = 106.867~g/mol\) Using the moles of iron(III) hydroxide, we can find the total mass: \(mass_{Fe(OH)_3} = moles_{Fe(OH)_3} × molar~mass = 0.00897~mol × 106.867~g/mol = 0.958~g\) Therefore, the mass of iron(III) hydroxide added to the HCl solution is approximately \(0.958~g\).

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Most popular questions from this chapter

For a product to be called "vinegar," it must contain at least \(5.0 \%\) acetic acid, \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\), by mass. A \(10.00-\mathrm{g}\) sample of a "raspberry vinegar" is titrated with \(0.1250 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) and required \(37.50 \mathrm{~mL}\) for complete neutralization. Can the product be called a "vinegar"?

The percentage of sodium hydrogen carbonate, \(\mathrm{NaHCO}_{3}\), in a powder for stomach upsets is found by titrating with \(0.275 M\) hydrochloric acid. If \(15.5 \mathrm{~mL}\) of hydrochloric acid is required to react with \(0.500 \mathrm{~g}\) of the sample, what is the percentage of sodium hydrogen carbonate in the sample? The balanced equation for the reaction that takes place is $$ \mathrm{NaHCO}_{3}(s)+\mathrm{H}^{+}(a q) \longrightarrow \mathrm{Na}^{+}(a q)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O} $$

Identify the type of aqueous reaction using the symbols PPT for precipitation, SA/SB for strong acid-strong base, SA/WB for strong acid-weak base, WA/SB for weak acid-strong base, and NR for no reaction. (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}+\mathrm{HCl}\) (b) \(\mathrm{Ca}(\mathrm{OH})_{2}+\mathrm{HF}\) (c) \(\mathrm{Ca}(\mathrm{OH})_{2}+\mathrm{Na}_{3} \mathrm{PO}_{4}\) (d) \(\mathrm{Ag}_{2} \mathrm{SO}_{4}+\mathrm{BaCl}_{2}\) (e) \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}+\mathrm{NaCl}\)

Write balanced equations for the following reactions in basic solution. (a) \(\mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{N}_{2} \mathrm{H}_{4}(a q) \longrightarrow \mathrm{Ni}(s)+\mathrm{N}_{2}(g)\) (b) \(\mathrm{Fe}(\mathrm{OH})_{3}(s)+\mathrm{Cr}^{3+}(a q) \longrightarrow \mathrm{Fe}(\mathrm{OH})_{2}(s)+\mathrm{CrO}_{4}{ }^{2-}(a q)\) (c) \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{BrO}_{3}^{-}(a q) \longrightarrow \mathrm{MnO}_{2}(s)+\mathrm{BrO}_{4}^{-}(a q)\) (d) \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{IO}_{4}^{-}(a q) \longrightarrow \mathrm{IO}_{2}^{-}(a q)+\mathrm{O}_{2}(g)\)

I A student is given \(0.930 \mathrm{~g}\) of an unknown acid, which can be either oxalic acid, \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\), or citric acid, \(\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}\). To determine which acid she has, she titrates the unknown acid with \(0.615 \mathrm{M} \mathrm{NaOH}\). The equivalence point is reached when \(33.6 \mathrm{~mL}\) are added. What is the unknown acid?
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