Asked by xx
Version A formular:
3Zn(OH)2 + 2Zn(CO)3(s)----------------------------->5ZnO(s)+2CO2(g) +3H2O(g)
105.4mL of Co2(g) was generated with a pressure of 765.5mm Hg for the gas. The temperature of the apparatus was 25.0C. How much zinc oxide would be left in tube if version A was the correct formula?
3Zn(OH)2 + 2Zn(CO)3(s)----------------------------->5ZnO(s)+2CO2(g) +3H2O(g)
105.4mL of Co2(g) was generated with a pressure of 765.5mm Hg for the gas. The temperature of the apparatus was 25.0C. How much zinc oxide would be left in tube if version A was the correct formula?
Answers
Answered by
DrBob222
Use PV = nRT and solve for n = mols CO2.
Then mols CO2 x (5 mols ZnO/2 mols CO2) = mols ZnO.
Then convert mols ZnO to g. g = mols x molar mass.
I would be somewhat concerned that CO2 is soluble in H2O, you haven't made a correction for the vapor pressure of H2O and you're classifying H2O as a gas at 25C.
Then mols CO2 x (5 mols ZnO/2 mols CO2) = mols ZnO.
Then convert mols ZnO to g. g = mols x molar mass.
I would be somewhat concerned that CO2 is soluble in H2O, you haven't made a correction for the vapor pressure of H2O and you're classifying H2O as a gas at 25C.
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