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 \(155 \mathrm{mL}\) ) $$ 2 \mathrm{Na}(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow 2 \mathrm{NaOH}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g}) $$

Short Answer

Expert verified
Around 0.624 grams of Sodium is required to produce a 0.175 M NaOH solution in 155 mL.

Step by step solution

01

Calculate moles of NaOH

In order to determine the amount of Sodium, the amount of moles of sodium hydroxide in the solution should firstly be calculated. The molarity of a solution is calculated by the number of moles of solute per liter of solution (M = moles/L). In this case, M = 0.175M and V = 155 mL = 0.155 L. Apply the formula \(Molarity = \frac{moles}{volume}\) to calculate moles. So, \(moles = Molarity \times Volume = 0.175 M \times 0.155 L = 0.0271 moles\) of NaOH.
02

Apply the stoichiometry from chemical equation

According to the balanced chemical equation, 2 moles of Na produce 2 moles of NaOH. Therefore, the moles of Na needed to produce 0.0271 moles of NaOH is also 0.0271 moles.
03

Calculate mass of Sodium

We know the molar mass of Hydrogen (Na) is about 23 g/mol. Therefore, the amount of Sodium needed is \(mass = moles \times molar mass = 0.0271 mol \times 23 g/mol = 0.624 g\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Azobenzene, an intermediate in the manufacture of dyes, can be prepared from nitrobenzene by reaction with triethylene glycol in the presence of \(\mathrm{Zn}\) and KOH. In one reaction, 0.10 L of nitrobenzene \((d=1.20 \mathrm{g} / \mathrm{mL})\) and \(0.30 \mathrm{L}\) of triethylene glycol \((d=1.12 \mathrm{g} / \mathrm{mL})\) yields \(55 \mathrm{g}\) azobenzene. What are the (a) theoretical yield, (b) actual yield, and (c) percent yield of this reaction? $$ 2 \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NO}_{2}+4 \mathrm{C}_{6} \mathrm{H}_{14} \mathrm{O}_{4} \frac{\mathrm{Zn}}{\mathrm{KOH}} $$ nitrobenzene triethylene glycol $$ \left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{N}\right)_{2}+4 \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{4}+4 \mathrm{H}_{2} \mathrm{O} $$ azobenzene

A small piece of zinc is dissolved in \(50.00 \mathrm{mL}\) of \(1.035 \mathrm{M}\) HCl. At the conclusion of the reaction, the concentration of the \(50.00 \mathrm{mL}\) sample is redetermined and found to be \(0.812 \mathrm{M} \mathrm{HCl} .\) What must have been the mass of the piece of zinc that dissolved? $$\mathrm{Zn}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{ZnCl}_{2}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g})$$

An organic liquid is either methyl alcohol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) ethyl alcohol \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\right),\) or a mixture of the two. A 0.220-g sample of the liquid is burned in an excess of \(\mathrm{O}_{2}(\mathrm{g})\) and yields \(0.352 \mathrm{g} \mathrm{CO}_{2}(\mathrm{g}) .\) Is the liquid a pure alcohol or a mixture of the two?

The incomplete combustion of gasoline produces \(\mathrm{CO}(\mathrm{g})\) as well as \(\mathrm{CO}_{2}(\mathrm{g}) .\) Write an equation for \((\mathrm{a})\) the complete combustion of the gasoline component octane, \(\mathrm{C}_{8} \mathrm{H}_{18}(\mathrm{l}),\) and \((\mathrm{b})\) incomplete combustion of octane with \(25 \%\) of the carbon appearing as \(\mathrm{CO}(\mathrm{g})\)

What volume of \(0.0175 \mathrm{M} \mathrm{CH}_{3} \mathrm{OH}\) must be added to \(50.0 \mathrm{mL}\) of \(0.0248 \mathrm{M} \mathrm{CH}_{3} \mathrm{OH}\) so that the resulting solution has a molarity of exactly \(0.0200 \mathrm{M}\) ? Assume that the volumes are additive.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free