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Chapter 4: Problem 61

A 0.3126 g sample of oxalic acid, \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4},\) requires 26.21 mL of a particular concentration of \(\mathrm{NaOH}(\mathrm{aq})\) to complete the following reaction. What is the molarity of the \(\mathrm{NaOH}(\mathrm{aq}) ?\) \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(\mathrm{s})+2 \mathrm{NaOH}(\mathrm{aq}) \longrightarrow\) $$ \mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) $$

Short Answer

Expert verified
The molarity of the \(\mathrm{NaOH(aq)}\) is 0.265 M.

Step by step solution

01

Convert the mass of oxalic acid to moles.

Using the molar mass of oxalic acid, which is 90.03g/mol, the number of moles of oxalic acid can be calculated as follows: Moles of oxalic acid = mass / molar mass = 0.3126g / 90.03g/mol = 0.00347 mol.
02

Apply stoichiometry.

According to the balanced chemical equation, 1 mol of oxalic acid reacts completely with 2 moles of \(NaOH\). Therefore, the moles of \(NaOH\) required for the complete reaction is double the moles of oxalic acid, i.e., = 2 * 0.00347 mol = 0.00694 mol.
03

Calculate molarity

The volume of \(NaOH\) solution used is given as 26.21 mL, which should be converted to liters as molarity is expressed in mol/L: Volume = 26.21 mL = 0.02621 L. From the definition of molarity (moles/volume), we can now calculate the molarity of the \(NaOH\) solution: Molarity of \(NaOH\) = moles/volume = 0.00694 mol / 0.02621 L = 0.265 M.

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Most popular questions from this chapter

Briefly describe (a) balancing a chemical equation; (b) preparing a solution by dilution; (c) determining the limiting reactant in a reaction.

Chlorine can be generated by heating together calcium hypochlorite and hydrochloric acid. Calcium chloride and water are also formed. (a) If \(50.0 \mathrm{g}\) \(\mathrm{Ca}(\mathrm{OCl})_{2}\) and \(275 \mathrm{mL}\) of \(6.00 \mathrm{M} \mathrm{HCl}\) are allowed to react, how many grams of chlorine gas will form? (b) Which reactant, \(\mathrm{Ca}(\mathrm{OCl})_{2}\) or \(\mathrm{HCl}\), remains in excess, and in what mass?

How many grams of sodium must react with \(155 \mathrm{mL}\) \(\mathrm{H}_{2} \mathrm{O}\) to produce a solution that is \(0.175 \mathrm{M} \mathrm{NaOH} ?\) (Assume a final solution volume of \(155 \mathrm{mL}\) ) $$ 2 \mathrm{Na}(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow 2 \mathrm{NaOH}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g}) $$

A \(25.0 \mathrm{mL}\) sample of \(\mathrm{HCl}(\mathrm{aq})\) is diluted to a volume of 500.0 mL. If the concentration of the diluted solution is found to be \(0.085 \mathrm{M} \mathrm{HCl}\), what was the concentration of the original solution?

A commercial method of manufacturing hydrogen involves the reaction of iron and steam. $$ 3 \mathrm{Fe}(\mathrm{s})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \stackrel{\Delta}{\longrightarrow} \mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{s})+4 \mathrm{H}_{2}(\mathrm{g}) $$ (a) How many grams of \(\mathrm{H}_{2}\) can be produced from \(42.7 \mathrm{g}\) Fe and an excess of \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (steam)? (b) How many grams of \(\mathrm{H}_{2} \mathrm{O}\) are consumed in the conversion of \(63.5 \mathrm{g}\) Fe to \(\mathrm{Fe}_{3} \mathrm{O}_{4} ?\) (c) If \(14.8 \mathrm{g} \mathrm{H}_{2}\) is produced, how many grams of \(\mathrm{Fe}_{3} \mathrm{O}_{4}\) must also be produced?
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