In Part A, you will produce a precpitate of calcium iodate by reacting aqueous solutions of potassium iodate and calcium nitrate.Ca(NO3)2 (aq) + 2 KIO3 (aq) → Ca(IO3)2 (s) + 2 KNO3(aq)You will react 10 mL of 1 mol L-¹ calcium nitrate with 25 mL of 0.2 mol L-¹ potassium iodate. What is the percentage yield of the reactionif 0.76 g of calcium iodate is produced?Hints:● Consider the stoichiometry of the reaction;● Calculate the number of moles of each of the starting materials used;This link shows an example of a yield calculation.Unrounded valueRounded valueNumber moles of potassium iodate used, (mol)0.005Number of moles of calcium nitrate used, (mol)0.01Molar Mass of calcium iodate, (g/mol)389.99Minimize QUESTION 1INTRODUCTIONCalculate the yield of calcium iodate produced in Part A from calcium nitrate solution and potassium iodate solution.Remember that you have to consider the stoichiometry of the reaction.Ca(NO3)2 (aq) + 2 KIO3 (aq) → Ca(IO3)2 (s) + 2 KNO3(aq)REAGENT DATACalcium nitratePotassium Iodate0.988Concentration, (mol L-¹)Volume, (mL)Number of moles, (mol)A424

Given following balanced Chemical equation - Ca(NO3)2(aq) + 2KIO3(aq) ----> Ca(IO3)2(s) +…

QUESTION 1 INTRODUCTION Calculate the yield of calcium iodate produced in Part A from calcium nitrate solution and potassium iodate solution. Remember that you have to consider the stoichiometry of the reaction. Ca(NO3)2 (aq) + 2 KIO3 (aq) → Ca(IO3)2 (s) + 2 KNO3(aq) REAGENT DATA Calcium nitrate Potassium Iodate 0.988 Concentration, (mol L-¹) Volume, (ml) Number of moles, (mol) 0

QUESTION 1 INTRODUCTION Calculate the yield of calcium iodate produced in Part A from calcium nitrate solution and potassium iodate solution. Remember that you have to consider the stoichiometry of the reaction. Ca(NO3)2 (aq) + 2 KIO3 (aq) → Ca(IO3)2 (s) + 2 KNO3(aq) REAGENT DATA Calcium nitrate Potassium Iodate 0.988 Concentration, (mol L-¹) Volume, (ml) Number of moles, (mol) 0

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

3. Given Molecular Equation for reaction of aqueous ammonia with acetic acid solution, based on observed demonstration indicate if the reaction is Electrolyte or non-electrolyte. Indicate/write all species including mobile ions, liquids, and solids present in product mixture. Write the Total ionic and net ionic equation. Molecular Equation: NH3(aq) + HC2H302(aq) → NH4C2H302(aq) Reaction mixture is Acetic Acid - HC2H302 NH4C2H302 Product mixture is Ammonium Acetate -
Balance the reaction, do not leave any fractions, don't leave anything blank, then answer the questions about the reaction ______ HNO3 + _______ Al → ________ H2(g) + ________ Al(NO3)3 This is an _____ reaction. Is this a redox reaction? (yes/no) ________, because Al0 is oxidized to Al+3, and because H+1 is _______ to H20. Will the reaction happen as it is written? (yes/no) _________ , because H2 is a gas and will leave the reaction container. The reverse reaction would not happen.
Show the net ionic equation when you mix potassium sulfate (K 2 SO 4 ) with barium nitrate (Ba(NO 3) 2 ) to create a solid barium sulfate precipitate (BaSO 4 ) .
QUESTION 22 A student titrates 55.2 mL of a sulfuric acid solution which contains 1.40 M NaOH. The color change indicates that it takes exactly 37.4 mL of the NaOH solution for the reaction to go to completion. What was the concentration of the original H₂SO4 solution? (Write your answer in 3 significant figures)? The balanced equation is: H2SO4 (aq) +2 NaOH(aq) → 2 H20 (1) + Na2SO4 U (atomic mass 235.0439) is 45.84% and the QUESTION 23 A sample of waste from a nuclear reactor contains an unnatural distribution of isotopes of uranium. The abundance of 235 ( abundance of 238 (atomic mass 238.0508) is the remainder. What would the average atomic mass of uranium be in this sample? (Write your answer in 4 significant figures. Do not include units.)
Complete and balance the equation for the reaction, where the reactants are in aqueous solution. ? Na₃PO₄ + ? Ba(NO₃)₂ → ? + ? Use no fractional coefficients. The number of moles and formula of the product containing Ba are A) Ba(PO₄)₂ B) BaPO₄ C) 3NaNO₃ D) Ba₃(PO₄)₂
Select all true statements from the following. Question 1 options: In order to make 600 mL of a 0.40 M solution of sodium hydroxide, it would be appropriate to put the required amount of solid sodium hydroxide in a container, and to then add 600 mL of water. If nitrogen reacts with fluorine to produce nitrogen trifluoride according to the equation N2 + 3 F2 → 2 NF3, fluorine is the limiting reagent if 7 moles of nitrogen and 11 moles of fluorine are mixed and allowed to react to completion. Na2SO4 is an acid. If salt is mixed with pepper, the result is a solution.
In the following reaction, 35.0 mL of 0.150 M magnesium bromide solution reacts with excess silver nitrate to form a precipitate. What is the expected mass of AgBr in grams? (molar mass of AgBr = 187.77 g/mole) MgBr2(aq) + 2AgNO3(aq) --> 2AgBr(s) + Mg(NO3)2(aq) Question 11 options: 0.985 g 0.493 g 1.97 g 986 g none of these
Question 14 of 15 Submit What mass of precipitate (in g) is formed when 20.5 mL of 0.600 M Cd(NO32), reacts with 26.0 mL of 0.600 M NaOH 3/2 in the following chemical reaction? Cd(NO3),(aq) + 2 NaOH(aq) – Cd(OH),(s) + 2 NaNO,(aq) g 1 2 3 4 C 7 8 9 +/- х 100 II |X
12. Sulfuric acid can dissolve aluminum metal according to the following reaction. 2 Al(s) + 3 H₂SO4(aq) Al2(SO4)3(aq)+3 H2(g) Suppose you wanted to dissolve an aluminum block with a mass of 29.0 g and you have 1.80 L of 1.00 M sulfuric acid available. a) Determine the limiting reactant. What is the theoretical yield of aluminum sulfate? 2/3/22 b) After the reaction is complete, you weigh the aluminum sulfate and it has a mass of 145 g. What is the percent yield of aluminum sulfate?
A sample of 0.75 g of hydrogen gas was obtained by reacting 32.5 g hydrochloric acid with 18.0 g magnesium. Mg(s) + 2 HCI (aq) --> MgCl₂ (aq) + H₂(g) What is the percent yield for this reaction? 100% 83% 41% 50%