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Volume of a Gas from a thermal decomposition
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[QUOTE="DottZakapa, post: 6453093, member: 559065"] [B]Homework Statement:[/B] Thermal decomposition of calcium hydrogen carbonate yields calcium carbonate water and carbon dioxide. Calculate the volume of carbon dioxide measured at 0 °C and 1.00 atm that is obtained by heating 80.0 grams of calcium carbonate [B]Relevant Equations:[/B] gas laws Ca(HCO[SUB]3[/SUB])2 -> CaCO[SUB]3[/SUB] + H[SUB]2[/SUB]O + CO[SUB]2[/SUB] First I evaluate the moles of calcium carbonate (don't mind the units, just to save time) ##\frac {80.0}{40,00+12.01+3*16,00}= 0,799 mol## From the equation, correct me if I am wrong , one mole of CaCO[SUB]3[/SUB] is proportional to one mole of CO[SUB]2[/SUB], so from this I can say that also CO[SUB]2[/SUB] has 0,799 mol. Using the ideal gas law equation PV=nRT, I can compute V: ##V=\frac{nRT}{P} ## inserting the values, I don't get the expected result, so I suspect that something is wrong, Any help ? [/QUOTE]
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Volume of a Gas from a thermal decomposition
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