Please help using the information below calculate the limiting reagent and the theoretical yield Freedom10:31 PMWe have to find the mass used in each reactant to create stock solution.Reactant:-1Potassium carbonate - [0.200M]Molar mass 147.01 gm/molReactant-2calcium chloride dihydrate= [0.220 M]Molay mass = 138.21 gm/mol.And we make 20 m² of each.Step2b)Solution :-Reactant-1 Potassiuma casi bonateConcentration (c) = 0·200 M = 0·200 mol/LVolume (V) = 20 mL = 20 L = 0·02 L [1L = 1000 m²]1000Now,Moles(n)Concentration (e) =Volume (V)Moles (n) = Concentration (C) x Volume (V) = 0-200 mol x 0·02 L = 0·004 mol.Molar mass = 147.01 gm/mol.Now.Mass = Molex Molar mass = 0.004 molx 147.01 gm = 0.588 gm.Hence, 0.588 gm of potassium carbonate is used.Reactant-2- Calcium chloride dihydrateConcentration (e) = 0·220 M = 0·220 mol/LVolume (v) = 20 mL = 0.02 LNow,c = nn = cx V = 0·220 mol x0·02 L = 0.0044 mol.Molar mass = 138.21 g/mol.NOW,molMass = Molex Molar mass = 0.0044 mdx138.21 m² = 0.608 gm.Hence, 0.608 gm of calcium chloride dihydrate is used.√xDo8

Limiting reagents in a chemical reaction is a reagents which completely consume in the reaction or…

ll Freedom 10:31 PM We have to find the mass used in each reactant to create stock solution. Reactant :-1 Potassium carbonate - [0.200M] Molar mass 147.01 gm/mol Reactant-2 calcium chloride dihydrate= [0.220 M] Molay mass=138-21 gm/mat. And we make 20 m² of each. Solution :- Reactant-1- Potassium carbonate Concentration (c) = 0·200 M = 0.200 mol/L Volume (V) = 20 mL = _20_ _ L = 0·02 L [1L = 1000 mL] 1000 Now, Moles(n) Concentration (e) = Volume (V) Moles (n) = Concentration (e) x Volume (V) = 0.200 m³¹ x 0·02 L = 0·004 mol. Molar mass = 147.01 gm/mol. Now. Mass = Mole x Molar mass = 0.004 molx 147.01 gm² = 0.588 gm. Hence, 0.588 gm of potassium carbonate is used. Reactant-2- Calcium chloride dihydrate Concentration (e) = 0·220 M = 0·220 mol/L Volume (v) = 20 mL = 0.02 L Now, c = n ⇒ n = cx V = 0·220 mel x0·02 L = 0·0044 mol. Molay mass = 138.21 gm/mol. NOW, mol Mass = Molex Molar mass = 0·0044 mdx138.21 gm² = 0.608 gm. Hence, 0.608 gm of calcium chloride dihydrate is used. ? Step2 b) Do 8

ll Freedom 10:31 PM We have to find the mass used in each reactant to create stock solution. Reactant :-1 Potassium carbonate - [0.200M] Molar mass 147.01 gm/mol Reactant-2 calcium chloride dihydrate= [0.220 M] Molay mass=138-21 gm/mat. And we make 20 m² of each. Solution :- Reactant-1- Potassium carbonate Concentration (c) = 0·200 M = 0.200 mol/L Volume (V) = 20 mL = _20_ _ L = 0·02 L [1L = 1000 mL] 1000 Now, Moles(n) Concentration (e) = Volume (V) Moles (n) = Concentration (e) x Volume (V) = 0.200 m³¹ x 0·02 L = 0·004 mol. Molar mass = 147.01 gm/mol. Now. Mass = Mole x Molar mass = 0.004 molx 147.01 gm² = 0.588 gm. Hence, 0.588 gm of potassium carbonate is used. Reactant-2- Calcium chloride dihydrate Concentration (e) = 0·220 M = 0·220 mol/L Volume (v) = 20 mL = 0.02 L Now, c = n ⇒ n = cx V = 0·220 mel x0·02 L = 0·0044 mol. Molay mass = 138.21 gm/mol. NOW, mol Mass = Molex Molar mass = 0·0044 mdx138.21 gm² = 0.608 gm. Hence, 0.608 gm of calcium chloride dihydrate is used. ? Step2 b) Do 8

Chemistry & Chemical Reactivity

10th Edition

ISBN:9781337399074

Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel

Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel

Chapter1: Basic Concepts Of Chemistry

Section: Chapter Questions

Problem 49RGQ: Fluoridation of city water supplies has been practiced in the United States for several decades. It...

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