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PLZ HELP...The decomposition reaction of calcium carbonate is represented by the following balanced equation :? CaCO3(s) ===> heat CaO(s) + CO2 (g) A 15.8-g sample of calcium carbonate (CaCO3) was heated in an open container to cause decomposition.Calculate the theoretical yield of CO2 expected to be produced according to the following equation: # of gram??

SciMann replied: "_____Decomposition (1 mole ---> 1 mole) g CO2 = MW CO2 g/mole CO2 * (1 mole CO2/mole CaCO3) * (15.8g CaCO3 / MW CaCO3 g/mole CaCO3) = ?? Plug and SOLVE Basic mathematics is a prerequisite to chemistry – I just try to help you with the methodology of solving the problem."

How to calculate the theoretical yield of this problem? This is the problem: The decomposition reaction of calcium carbonate is represented by the following balanced equation: CaCO3(s) => CaO(s) + CO2(g) A 15.8-g sample of calcium carbonate (CaCO3) was heated in an open container to cause decomposition. Calculate the theoretical yield of CO2 expected to be produced according to the following equation: can someone please help me going step by step with this problem? thanx!

Jennifer R replied: "You need to determine the number of moles of Calcium Carbonate used. Because your equation is balanced at a 1:1 ratio for both the reactant and the products, the yield of CO2 should be the same # of moles as the moles of Calcium Carbonate. Calcium Carbonate has a molar mass of 100.087 g/mol. So, dividing your 15.8g sample by 100.087 gives the number of CaCO3 moles, which is 15.8/100.087 = 0.15786 moles of Calcium carbonate. You know you'll have the same number of moles of CO2, so your products will be 0.15786 moles of CaO and 0.15786 moles of CO2. In order to figure out the theoretical yield, multiply the number of moles of CO2 by the mass of a mole of CO2(44.0096 g), and you have your answer. (0.15786x44.0096)= 6.94735546 g of CO2.. ..Tried to break it down as much as possible. Good luck!"

REGNUM replied: "Theoretical yield: (grams A) x (Molar Mass A) x (Mole Ratio) x (Molar Mass B) = grams of B CaCO3(s) => CaO(s) + CO2(g) 15.8g CaCO3 x (1mol CaCO3 / 100.0872g CaCO3) x (1mol CO2/ 1mol CaCO3) x (44.0098g CO2/ 1mol CO2) = 6.947490189 = 6.95g CO2 If you have any questions about anything done here let me know and ill answer that for you."

James replied: "15.8 g of CaCO3 (m.w. = 40+12+3*16=100 g) is 0.158 moles so you should leave behind 0.158 moles of CaO (m.w.=40+16 = 56 g) = 8.85 g and generate 0.158 moles of CO2 (m.w. = 12+2*16 = 44 g) = 6.95 g but without the equation we can't figure out more than that."

vij replied: "CaCO3(s) => CaO(s) + CO2(g) Substances react in the ratio of moles as in the balanced chemical equation. The given equation tells that one mole of calcium carbonate on heating gives one mole of Carbon dioxide. no .of moles of CaCO3 in 15.8 g = 15.8 g/ molar mass = 15.8/100g/mole = .158 mole moles of CO2 produced = 1mole of CO2/1mole of CaCO3 X 0.158 mole of CaCO3 = 0.158 mole Amt of CO2 = 0.158 mole x 44 g/mole = 6.952 g"

Chemistry question asking: What is the final pressure (in atm) inside the container? Calcium carbonate decomposes upon heating to form calcium oxide and carbon dioxide. A 1.75g sample of calcium carbonate is placed in an empty, sealed 2.00L steel container. The container is heated to 625C and then allow to cool to 25C. What is the final pressure (in atm) inside the container? R = 0.0821 L atm mol^-1 K^-1 please show work! thanks so much!

Dr. Buzz replied: "Allie, First we need to write the balanced equation: CaCO3 ---> CaO + CO2 -- a 1:1 correspondence in moles Moles CO2 = moles CaCO3 = 1.75 g/(100 g/mole CaCO3) = 1.75 x 10^-2 moles Pressure = nRT/V = (1.75 x 10^-2 moles)*(0.0821 L*atm/mole*K)*(273.15 + 25)/(2.00 L) = 0.214 atm Hope that helped!"

Mildred replied: "First you have to find out how much calcium caubonate weighs: CaCO3 Ca= 40.078 C= 12.0107 O= 15.9994 40.078+12.0107+(3*15.9994) =100.0869 Then you have to turn the 1.75 grams into moles using the atomic weight you just figured out: 1.75g sample/100.0869 atomic weight = .01748 moles Then you just plug in what you know into PV=nRT P(2)=(.01748)(.0821)(25) 2P=.0358777 P=.0179 atm"

The decomposition reaction of calcium carbonate is represented by the following balanced equation:? CaCO3(s) ---Heat---> CaO(s) + CO2(g) A 15.8 Gram sample of CaCO3 was heated in an open container to cause decomposition. calculate the theoretical yield of CO2 expected to be produced according to the following equation? Answer must be in grams.

Flyboy replied: "Here's how you do this the easy way: Add the molecular weights of CaO and CO2. Divide the molecular weight of CO2 by the sum above and multiply this fraction by the starting amount, 15.8 grams. Surely, a future Chemistry Nobel Prize winner like yourself can take it from here."

RONALD E B replied: "1 mole of CaCO3 yields 1 mole of CO2 15.8 g CaCO3 = 0.159 moles of CaCO3 0.159 moles of CO2 = 7.00 grams of CO2 convert grams to moles equate moles to moles (1:1 in this case) convert moles to grams"

papastolte replied: "Find mol CaCO3. See from the balanced equation that 1 mol CaCO3 ~ 1 mol CO2. Find mol CO2. Find g CO2."

predict and balance the following synthesis and decomposition reactions.? a) A sample of calcium carbonate is heated. b) Sulfur Dioxide gas is bubbled through water. c) Solid potassium oxide is added to a container of carbon dioxide gas. d) liquid hydrogen peroxide is warmed. e) Solid lithium oxide is added to water. f) molten aluminum chloride is electrolyzed. g) A pea sized piece of sodium is added to a container of iodine vapor. h) A sample of carbonic acid is heated. i) A sample of potassium chlorate is heated. j) Solid magnesium oxide is added to sulfur trioxide gas.

ninjamayn replied: "a) CaCO3 ~~> CaO + CO2 (Decomposition reactions usually produce simple salts and oxide gasses) b) SO2 + H2O ~~> H2SO3 (A nonmetal oxide in water will produce an acid) c) K2O + CO2 ~~> K2CO3 (CO2 + O ~~> CO3(2-)) d) 2 H2O2 ~~> 2 H2O + O2 (Decomposition reaction) e) Li2O + H2O ~~> 2(Li+) + 2(OH-) (A metal oxide in water will produce a base) f) 2 AlCl3 ~~> 2 Al(3+) + 3 Cl2 + 3e- (3 electrons are taken away from Al giving it its 3+ charge. The 3 electrons are given to the 3 Cl(1-) which converts them into a more stable form Cl2 gas. Al was oxidized and Cl was reduced.) g) 2Na + I2 ~~> 2 NaI (Na is oxidized and I is reduced) h) H2CO3 ~~> CO2 + H2O (Carbonic acid always decomposes into water and carbon dioxide) i) 2 KClO3 ~~> 2 KCl + 3 O2 (Decomposition reactions) j) MgO + SO3 ~~> MgSO4 (SO3 + O ~~> SO4(2-))"

how to find Percent yield and theoretical yield? When heated, calcium carbonate undergoes a decomposition reaction represented by the following chemical equation. CaCO3(s) CaO(s) + CO2(g) After a student heated a 15.8-g sample of calcium carbonate in an open container to cause decomposition, the mass of the remaining solid was measured to be 9.10 g. In order to determine if the reaction was complete or not, she calculated the following information. The theoretical yield of CO2 =___________ g The percent yield of CO2 in her process =__________%

SciMann replied: "_____Lab yield ________CaCO3(s) -----------------------------> CaO(s) + CO2(g) ________1 mole----------------------------------> 1 mole__1mole _init_____x=15.8g/MW CaCO3 g/mole_____0_______0 _theory___0___________________________x_______x _exp_____x-y__________________________y_______y Since NO CO2 dat is given, NO answers can be given with regard to CO2, except by inference from the amount of CaO generated. theoretical yield of CO2 = x moles * MW CO2 g/mole = 15.8g/MW CaCo3 g/mole * MW CO2 g/mole = ?? g CO2 y # moles, if all of product is CaO = 9.10 g / MW CaO = ?? mole "CaO" expt x # moles CaO = 15.8 g / MW CaCO3 = ?? max moles of "CaO" possible since ?? moles "CaO" expt > ?? max moles of "CaO", the reaction is not 100% and some of the residual mass is unreacted CaCO3 residue g = moles CaCO3 * MW CaCO3 g/mole + moles CaO * MW Cao g/mole moles CaCO3 = (x - y) moles; x is known moles CaO = y moles residue g = (x - y) * MW CaCO3 g/mole + y * MW CaO g/mole ________= x mole * MW CaCO3 g/mole + y * (MW CaO - MW CaCO3) g/mole Solve for y expt % yield = 100% * y/x = ?? % Plug and SOLVE (a little over 95% yield) Basic mathematics is a prerequisite to chemistry – I just try to help you with the methodology of solving the problem."

please help me solve this chemistry problem! CaCO3 ---> CaO + CO2? After 15.8 g of sample of calcium carbonate was heated in an open container to cause decomposition, the mass of the remaining solid was determined to be 9.10 g. Calculate the mass of the carbon dioxide produced. ..This is a question in my chem book, and I have the answer but can someone show me the steps because I cannot come up with the answer in the book. Thank you!

lisa replied: "use the theory of conservation of mass. this states that mass cannot be lost or gained durring a reaction meaning that if u start off with a sample of 15.8g, than your products should weight the same as the CaCO3. just subtract the mass of CaO from the total mass u began with(15.8) go get the mass of CO2 gas created"

What is the percentage of silica in sandstone? A sample of sandstone consists of silica, SiO2, and calcite, CaCO3. When the sandstone is heated, calcium carbonate, CaCO3, decomposes into calcium oxide, CaO, and carbon dioxide CaCO3(s) --> CaO(s) + CO2(g) what is the percentage of silica in the sandstone if 23.5 mg of the rock yields 5.85 mg of carbon dioxide?

Timothy H replied: "CaCO3 has a molecular wt of 100 CO2 has a molecular wt of 44 Calculation: 5.85mg x 100 / 44 = 13.3mg CaCO3 100% x (23.5mg - 13.3mg) / 23.5mg = 43.4% Silica"

Urgent chem problem, please help. EASY!? Can I have a molecular and ionic equation for: A sample of calcium carbonate is heated.

haile d replied: "CaCO3 -------> CaO + CO2 CO32- ----> O2- + CO2"

8 ball replied: "CaCO3(s) → CaO(s) + CO2(g) no ionic equation because you have no aqueous solution, so no ions are formed."

Chemistry help please!!? After a 15.8g sample of calcium carbonate was heated in an open container to cause decomposition, the mass of the remaining solid was determined to be 9.10g. Calculate the mass of carbon dioxide. Thanks for any help you can give me on this question.

Dr.A replied: "CaCO3 >> CaO + CO2 mass CO2 = 15.8 - 9.10 = 6.7 g"

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