Problem Quote \"In an experiment, 1.056g of a metal carbonate, containing andunk

Question

Problem Quote "In an experiment, 1.056g of a metal carbonate, containing andunknown metal M, is heated to give the metal oxide and 0.376gCO2. MCO3(S) + heat -> MO(S) +CO2(g) What is the identity of the metal M? Ni, Zn, Cu, or Ba" Given the masses, I know that MO = .68g, but that is about asfar as I can get. I have a few ideas of what I should do but nonethat actually solve it. Any help pointing in the right direction orshowing me how to solve this would be appreciated and ratedappropriately. Problem Quote "In an experiment, 1.056g of a metal carbonate, containing andunknown metal M, is heated to give the metal oxide and 0.376gCO2. MCO3(S) + heat -> MO(S) +CO2(g) What is the identity of the metal M? Ni, Zn, Cu, or Ba" Given the masses, I know that MO = .68g, but that is about asfar as I can get. I have a few ideas of what I should do but nonethat actually solve it. Any help pointing in the right direction orshowing me how to solve this would be appreciated and ratedappropriately. What is the identity of the metal M? Ni, Zn, Cu, or Ba" Given the masses, I know that MO = .68g, but that is about asfar as I can get. I have a few ideas of what I should do but nonethat actually solve it. Any help pointing in the right direction orshowing me how to solve this would be appreciated and ratedappropriately.

Explanation / Answer

We Know that :       The given equation is :         MCO3(S)+ heat -> MO(S) + CO2(g)       weight of theMCO3 = 1.056 g        weight of the metal oxide= 0.68 g         weight of theCO2 = 0.376 g           numberof moles of CO2 = 0.376 g / 44 g /mol                                                   = 0.008545 mol           From the balanced equation it is clear that 1 mole ofMCO3 gives 1 mole of MO and 1 mole ofCO2             number of moles of MCO3 = 0.008545mol = number of metal carbonate = number of molesof MO          0.008545= 1.056 g / M + 12 + 48           M = 63.57 g / mol          similarlyfrom the number of moles of MO we can obtain the weight ofthe metal          0.008545 = 0.68 / M + 16           M = 63.57 g / mol           From the inspection of the weight of the metal it is Cu                           number of moles of MCO3 = 0.008545mol = number of metal carbonate = number of molesof MO          0.008545= 1.056 g / M + 12 + 48           M = 63.57 g / mol          similarlyfrom the number of moles of MO we can obtain the weight ofthe metal          0.008545 = 0.68 / M + 16           M = 63.57 g / mol           From the inspection of the weight of the metal it is Cu              

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