Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Some sulfuric acid is spilled on a lab bench. You can neutralize the acid by sprinkling sodium bicarbonate on it and then mopping up the resulting solution. The sodium bicarbonate reacts with sulfuric acid according to: \begin{equation} \begin{array}{r}{2 \mathrm{NaHCO}_{3}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+} \quad\\\ {2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{CO}_{2}(g)}\end{array} \end{equation} Sodium bicarbonate is added until the fizzing due to the formation of \(\mathrm{CO}_{2}(g)\) stops. If 27 \(\mathrm{mL}\) of 6.0 \(\mathrm{MH}_{2} \mathrm{SO}_{4}\) was spilled, what is the minimum mass of \(\mathrm{NaHCO}_{3}\) that must be added to the spill to neutralize the acid?

Short Answer

Expert verified
The minimum mass of NaHCO3 needed to neutralize the spill of 27 mL of 6.0 M H2SO4 is approximately 27.22 g.

Step by step solution

01

Find the moles of H2SO4

First, we need to calculate the moles of H2SO4 using the given volume (27 mL) and molarity (6.0 M). The formula to calculate moles is: Moles = Molarity × Volume (in Liters) So, moles of H2SO4 = 6.0 M × (27 mL × 0.001 L/mL) = 6.0 M × 0.027 L = 0.162 moles of H2SO4
02

Find the moles of NaHCO3 needed

Now that we have the moles of H2SO4, we'll use the balanced chemical equation to find the corresponding moles of NaHCO3 needed for neutralization: 2 NaHCO3(s) + H2SO4(aq) → Na2SO4(aq) + 2 H2O(l) + 2 CO2(g) From the balanced equation, it's clear that 2 moles of NaHCO3 are required to neutralize 1 mole of H2SO4. So, the moles of NaHCO3 required = 2 × Moles of H2SO4 = 2 × 0.162 = 0.324 moles of NaHCO3
03

Convert moles of NaHCO3 to mass

Finally, we'll convert the moles of NaHCO3 to mass using the molar mass of NaHCO3. We can find the molar mass by adding the atomic masses of each element in the compound: Molar mass of NaHCO3 = (22.99 g/mol Na) + (1.008 g/mol H) + (12.01 g/mol C) + (3 × 16.00 g/mol O) = 84.007 g/mol Now, we can convert moles of NaHCO3 to mass: Mass of NaHCO3 = Moles × Molar mass = 0.324 moles × 84.007 g/mol = 27.22 g (approximately) So, the minimum mass of NaHCO3 needed to neutralize the spill is approximately 27.22 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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molarity
Molarity is an essential concept in chemistry that represents the concentration of a solute in a solution. It indicates how many moles of a substance are present in one liter of solution, and is expressed in units of moles per liter (M).
When you know the molarity of a solution, you can calculate the amount of solute in any given volume of that solution. This makes molarity a crucial concept for converting between the volume of a liquid and the quantity of solute it contains, especially in reactions like neutralization.
  • Molarity formula: \( M = \frac{moles \text{ of solute}}{volume \text{ in liters}} \)
  • For example, a 6.0 M sulfuric acid solution means there are 6 moles of sulfuric acid per liter of solution.
Moles Calculation
Calculating moles is a fundamental skill in chemistry that involves matching quantities of substances to their chemical formulas. It helps determine the amount of a chemical involved in a reaction.
Using molarity, you can calculate moles by multiplying the solution's molarity by its volume in liters.
For instance, to find the moles of sulfuric acid in 27 mL of a 6.0 M solution:
  • Convert the volume from mL to L: 27 mL \( \times \) 0.001 = 0.027 L
  • Multiply the molarity by the volume: 6.0 M \( \times \) 0.027 L = 0.162 moles of \( \text{H}_2\text{SO}_4 \)
This calculation is key to knowing the amounts needed for chemical reactions.
Balanced Chemical Equation
A balanced chemical equation is like a recipe in chemistry which ensures that the same number of atoms for each element is represented on both sides of the equation.
In our neutralization reaction:
  • \( 2 \text{NaHCO}_3 (s) + \text{H}_2\text{SO}_4 (aq) \rightarrow \text{Na}_2\text{SO}_4 (aq) + 2 \text{H}_2\text{O} (l) + 2 \text{CO}_2 (g) \)
  • This equation shows that two moles of \( \text{NaHCO}_3 \) react with one mole of \( \text{H}_2\text{SO}_4 \) to produce sodium sulfate and water, while releasing carbon dioxide gas.
Balanced equations are crucial for accurately predicting the outcome of chemical reactions.
Molar Mass
The molar mass is the mass of one mole of a substance. It is measured in grams per mole and helps in converting between the mass of a substance and the moles of that substance.
To calculate the molar mass of a compound, you add up the atomic masses of all the atoms in its formula.
  • For \( \text{NaHCO}_3 \):
  • Sodium (Na): 22.99 g/mol
  • Hydrogen (H): 1.008 g/mol
  • Carbon (C): 12.01 g/mol
  • Oxygen (O): 3 \( \times \) 16.00 g/mol
  • Total: 84.007 g/mol
This value is then used to convert from moles to grams in chemical calculations.
Chemical Stoichiometry
Chemical stoichiometry is the study of the quantitative relationships between the amounts of reactants and products in a chemical reaction.
It relies heavily on the use of balanced equations to determine how much of each substance is needed or produced.
In our neutralization reaction, for every mole of sulfuric acid, two moles of sodium bicarbonate are required, as shown by the stoichiometric coefficients in the equation.
  • Used to calculate: \( 2 \times 0.162 = 0.324 \) moles of \( \text{NaHCO}_3 \) needed.
  • This stoichiometric ratio ensures the proper amounts of substances are mixed to achieve complete neutralization.
Understanding stoichiometry is key to successfully conducting reactions in the lab.

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

State whether each of the following statements is true or false. Justify your answer in each case. \begin{equation} \begin{array}{l}{\text { (a) When methanol, } \mathrm{CH}_{3} \mathrm{OH}, \text { is dissolved in water, a con- }} \\ {\text { ducting solution results. }} \\ {\text { (b) When acetic acid, } \mathrm{CH}_{3} \mathrm{COOH} \text { , dissolves in water, the }} \\ {\text { solution is weakly conducting and acidic in nature. }}\end{array} \end{equation}

A 1.248 -g sample of limestone rock is pulverized and then treated with 30.00 mL of 1.035\(M\) HCl solution. The excess acid then requires 11.56 \(\mathrm{mL}\) of 1.010 \(\mathrm{M}\) NaOH for neutralization. Calculate the percentage by mass of calcium carbonate in the rock, assuming that it is the only substance reacting with the HCl solution.

(a) Which will have the highest concentration of potassium ion: \(0.20 M \mathrm{KCl}, 0.15 M \mathrm{K}_{2} \mathrm{CrO}_{4},\) or 0.080\(M \mathrm{K}_{3} \mathrm{PO}_{4} ?\) (b) Which will contain the greater number of moles of potassium ion: 30.0 \(\mathrm{mL}\) of 0.15 \(\mathrm{M} \mathrm{K}_{2} \mathrm{CrO}_{4}\) or 25.0 \(\mathrm{mL}\) of 0.080 \(\mathrm{M} \mathrm{K}_{3} \mathrm{PO}_{4} ?\)

Gold is isolated from rocks by reaction with aqueous cyanide, \(\mathrm{CN} : 4 \mathrm{Au}(s)+8 \mathrm{NaCN}(a q)+\mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) \(4 \mathrm{Na}\left[\mathrm{Au}(\mathrm{CN})_{2}\right](a q)+4 \mathrm{NaOH}(a q) .\) (a) \(\mathrm{Which}\) atoms from which compounds are being oxidized, and which atoms from which compounds are being reduced? (b) The \(\left[\mathrm{Au}(\mathrm{CN})_{2}\right]^{-}\) ion can be converted back to \(\mathrm{Au}(0)\) by reaction with \(\mathrm{Zn}(s)\) powder. Write a balanced chemical equation for this reaction. (c) How many liters of a 0.200\(M\) sodium cyanide solution would be needed to react with 40.0 \(\mathrm{kg}\) of rocks that contain 2.00\(\%\) by mass of gold?

Which ions remain in solution, unreacted, after each of the following pairs of solutions is mixed? \begin{equation} \begin{array}{l}{\text { (a) potassium carbonate and magnesium sulfate }} \\\ {\text { (b) lead nitrate and lithium sulfide }} \\ {\text { (c) ammonium phosphate and calcium chloride }}\end{array} \end{equation}

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