Solid silver can be produced by reacting silver (I) nitrate with copper according to the reaction below. What is the percent yield of a reaction if 9.876 g of Cu was placed in silver (I) nitrate solution and produced 13.147 g of silver (Assume the AGNO3 is in excess)? Cu(s) + 2 AGNO3(aq)→ Cu(NO3)2(aq) + 2 Ag(s) Your Answer: Answer units

Solid silver can be produced by reacting silver (I) nitrate with copper according to the reaction below. What is the percent yield of a reaction if 9.876 g of Cu was placed in silver (I) nitrate solution and produced 13.147 g of silver (Assume the AGNO3 is in excess)? Cu(s) + 2 AGNO3(aq)→ Cu(NO3)2(aq) + 2 Ag(s) Your Answer: Answer units

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

Upon reaction of 1.27 g of copper sulfate with excess zinc metal, 0.272 g copper metal was obtained as an experimental result according to the equation: CuSO4(aq) + Zn(s) → Cu(s) + ZnSO4(aq) Before the experiment was carried out the theoretical yield calculation was completed to predict the maximum amount of copper that could be isolated. The answer was 0.506 g of Cu. What is the percent yield in standard notation? [report your answer to the tenths place, i.e. one decimal place.]
How many grams of AgCI will be formed when 60 mL of 0.500 M AGNO3 is completely reacted according to the balanced chemical reaction: FeCl3(aq) + 3 AGNO3(aq) → 3 AgCI(s) + Fe(NO3)2(aq)
You are asked to create a 0.35 M solution of calcium chloride (CaCl2) using a 250.0 mL volumetric flask. a. You react 30.0 mL of 0.35 M CaCl2 with excess solid lithium. The balanced reaction is shown below. What mass of solid calcium is made during this experiment? CaCl2(aq) + 2 Li(s) → Ca(s) + 2 LiCl(aq)
What mass of precipitate (in g) is formed when 20.5 mL of 0.300 M Ni(NO₃)₂ reacts with 25.5 mL of 0.300 M NaOH in the following chemical reaction? Ni(NO₃)₂(aq) + 2 NaOH(aq) → Ni(OH)₂(s) + 2 NaNO₃(aq)
Aluminum nitrate reacts with sodium hydroxide according the the equation below. When 14.51 grams of aluminum nitrate is added to an excess amount sodium hydroxide, how many moles of aluminum hydroxide will be produced by the time the reaction is 47% complete? Report your answer with three significant figures. Al(NO3)3 (s) + 3 NaOH (aq) → Al(OH)3 (s) + 3 NaNO3 (aq)
Determine the mass, in grams, of calcium chloride that is formed when 39.6 mL of 0.437 M calcium hydroxide reacts with excess chloride gas? Please provide your answer to three (3) decimal places. The balanced equation is below. 2Ca(OH)2 (aq) + 2Cl2 (g) → Ca(OCI)2 (aq) + CaCl2 (s) + 2H₂O (1)
One can make a silver penny by plating silver onto older (pre-1982) copper pennies according to the following reaction. Cu(s) + 2 AgNO3(aq) 2 Ag(s) + Cu(NO3)2(aq) 1.27 grams of Copper were added to a solution that contains 5.10 grams of AgNO3. If only 1.6 grams of Silver were obtained, what was the % yield?
Sulfuric acid dissolves aluminum metal according to the following reaction: 2Al(s)+3H2SO4(aq)→Al2(SO4)3(aq)+3H2(g) Suppose you wanted to dissolve an aluminum block with a mass of 15.6 g . a) What minimum mass of H2SO4 would you need? Express your answer in grams. b)What mass of H2 gas would be produced by the complete reaction of the aluminum block?Express your answer in grams.
43.00 mL of 0.980 M barium hydroxide is added to 24.00 mL of 1.19 M aluminum chloride. Aluminum hydroxide precipitates. How many grams of aluminum hydroxide are produced? Clearly label the limiting and excess reactants. You must show all work including molar ratios. 2AlCl3(aq) + 3Ba(OH)2(aq) → 2Al(OH)3(s) + 2BaCl2(aq)
What mass of precipitate (in g) is formed when 20.5 mL of 0.500 M Cu(NO₃)₂ reacts with 29.0 mL of 0.500 M NaOH in the following chemical reaction? Cu(NO₃)₂(aq) + 2 NaOH(aq) → Cu(OH)₂(s) + 2 NaNO₃(aq)
Upon reaction of 1.27 g of copper sulfate with excess zinc metal, 0.314 g copper metal was obtained as an experimental result according to the equation: CuSO4(aq) + Zn(s) → Cu(s) + ZnSO4(aq) Before the experiment was carried out the theoretical yield calculation was completed to predict the maximum amount of copper that could be isolated. The answer was 0.506 g of Cu. What is the percent yield in standard notation?