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Which Equation Is Correct?

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Which Equation Is Correct? Aim: To design and plan an experiment to measure a volume of gas that will prove which of the following two equations is correct. Equation 1: 2CuCO3(s) � Cu2O(s) + 2CO2(g) + 1/2O2(g) Ratio of 4:5 Equation 2: CuCO3(s) � CuO(s) + CO2(g) Ratio of 1:1 A known mass of copper carbonate will be heated, and the volume of gas given off will be measured. From this, the number of moles of gas can be calculated. This will allow you to find out which equation is correct, because it is stated that each reaction produces a different number of moles of gas when heated. ...read more.


The aim is to collect about 90cm3 of gas. In standard conditions, the volume of one mole of gas is 24dm3. From this the amount gas each reaction will produce can be worked out, which allows you to work out how much CuCO3 to use. 90/24000 = 0.00375 moles of gas. As the reactant and products in equation 2 are in the same ratio, you can see that equation 1 will produce more gas as the ratio of that equation is 4:5. Ratio of 4:5 - 0.00300 : 0.00375 moles of gas. Therefore, to work out the mass of the CuCO3 the following equation should be used: Number of moles = Mass/Molecular Mass Mass = Number of moles x Molecular Mass 0.00300 x 123.5 = 0.3705g. ...read more.


* Heat the CuCO3 in the boiling tube until no more gas is released and the powder has become black. * Repeat twice for reliable and accurate results. Using the Results: The volumes of gas produced can be used to work out which equation is correct. To find the number of moles of gas given off, the volume needs to be divided by 24dm3. Number of moles = Volume (dm3) / 24 (dm3) Equation 1 - Ratio 4:5 - 0.00296 : 0.00375 x 24000 = 90cm3 Equation 2 - Ratio 1:1 - 0.00296 : 0.00296 x 24000 = 71.04cm3 This means that if 90cm3 of gas is produced, equation 1 is correct. However, if 71.04cm3 of gas is produced, equation 2 is correct. ...read more.

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