Concept explainers
(a)
Interpretation:
The volume of
(a)
Explanation of Solution
Given Information:
The given ionic reaction given is:
The molarity of
So, moles of solute can be calculated by rearranging the equation (1).
Moles of
The mole ratio of the reaction is:
Volume of
So,
(b)
Interpretation:
The mass of
(b)
Explanation of Solution
Given Information:
The given ionic reaction is:
To calculate the mass of
The molar mass of
So,
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Chapter 11 Solutions
INTO TO CHEMISTRY (EBOOK ACCESS CODE)
- Arsenic acid, H3AsO4, is a poisonous acid that has been used in the treatment of wood to prevent insect damage. Arsenic acid has three acidic protons. Say you take a 25.00-mL sample of arsenic acid and prepare it for titration with NaOH by adding 25.00 mL of water. The complete neutralization of this solution requires the addition of 53.07 mL of 0.6441 M NaOH solution. Write the balanced chemical reaction for the titration, and calculate the molarity of the arsenic acid sample.arrow_forward2. Equal amounts (moles) of acetic acid(aq) and sodium sulfite, Na2SO3(aq), are mixed. The resulting solution is acidic basic neutralarrow_forwardA 25.0-mL sample of sodium sulfate solution was analyzed by adding an excess of barium chloride solution to produce barium sulfate crystals, which were filtered from the solution. Na2SO4(aq)+BaCl2(aq)2NaCl(aq)+BaSO4(s) If 5.719 g of barium sulfate was obtained, what was the molarity of the original Na2SO4 solution?arrow_forward
- A soluble iodide was dissolved in water. Then an excess of silver nitrate, AgNO3, was added to precipitate all of the iodide ion as silver iodide, AgI. If 1.545 g of the soluble iodide gave 2.185 g of silver iodide, how many grams of iodine are in the sample of soluble iodide? What is the mass percentage of iodine, I, in the compound?arrow_forwardA student is asked to identify the metal nitrate present in an aqueous solution. The cation in the solution can be either Na+, Ba2+, Ag+, or Ni2+. Results of solubility experiments are as follows: unknown + chloride ions—no precipitate unknown + carbonate ions—precipitate unknown + sulfate ions—precipitate What is the cation in the solution?arrow_forwardFor a product to be called vinegar, it must contain at least 5.0% acetic acid, HC2H3O2, by mass. A 10.00-g sample of a raspberry vinegar is titrated with 0.1250 M Ba(OH)2 and required 37.50 mL for complete neutralization. Can the product be called a vinegar?arrow_forward
- A 0.608-g sample of fertilizer contained nitrogen as ammonium sulfate, (NH4)2SO4. It was analyzed for nitrogen by heating with sodium hydroxide. (NH4)2SO4(s)+2NaOH(aq)Na2SO4(aq)+2H2O(l)+2NH3(g) The ammonia was collected in 46.3 mL of 0.213 M HCl (hydrochloric acid), with which it reacted. NH3(g)+HCl(aq)NH4Cl(aq) This solution was titrated for excess hydrochloric acid with 44.3 mL of 0.128 M NaOH. NaOH(aq)+HCl(aq)NaCl(aq)+H2O(l) What is the percentage of nitrogen in the fertilizer?arrow_forwardA 8.50 g sample of KCl is dissolved in 66.0 mL of water. The resulting solution is then added to 72.0 mL of a 0.280 M CaCl2(aq) solution. Assuming that the volumes are additive, calculate the concentrations of each ion present in the final solution.arrow_forwardA 25.0-mL sample of vinegar (which contains the weak acid acetic acid, CH3CO2H) requires 28.33 mL of a 0.953 M solution of NaOH for titration to the equivalence point. What is the mass of acetic acid (molar mass = 60.05 g/mol), in grams, in the vinegar sample, and what is the concentration of acetic acid in the vinegar? CH3CO2H(aq) + NaOH(aq) NaCH3CO2(aq) + H2O(l)arrow_forward
- An antacid tablet contains sodium hydrogen carbonate, NaHCO3, and inert ingredients. A 0.465-g sample of powdered tablet was mixed with 53.3 mL of 0.190 M HCl (hydrochloric acid). The mixture was allowed to stand until it reacted. NaHCO3(s)+HCl(aq)NaCl(aq)+H2O(l)+CO2(g) The excess hydrochloric acid was titrated with 54.6 mL of 0.128 M NaOH (sodium hydroxide). HCl(aq)+NaOH(aq)NaCl(aq)+H2O(l) What is the percentage of sodium hydrogen carbonate in the antacid?arrow_forwardSodium chloride is used in intravenous solutions for medical applications. The NaCl concentration in such solutions must be accurately known and can be assessed by reacting the solution with an experimentally determined volume of AgNO3 solution of known concentration. The net ionic equation is Ag+(aq)+Cl(aq)AgCl(s) Suppose that a chemical technician uses 19.3 mL of 0.200-M AgNO3 to convert all the NaCl in a 25.0-mL sample of an intravenous solution to AgCl. Calculate the molarity of NaCl in the solution.arrow_forwardA metal, M, was converted to the chloride MCl2. Then a solution of the chloride was treated with silver nitrate to give silver chloride crystals, which were filtered from the solution. MCl2(aq)+2AgNO3(aq)M(NO3)2(aq)+2AgCl(s) If 2.434 g of the metal gave 7.964 g of silver chloride, what is the atomic weight of the metal? What is the metal?arrow_forward
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