Question

What is the volume of co2 obtained on heating 20g of 95% pure MgCO3

Answer 1

We address the stoichiometric equation…

MgCO3(s)+Δ→MgO(s)+CO2(g)↑

And this is the general decomposition reaction of carbonates…we assume that 10.5∙g magnesium carbonate, a SPECIFIC molar quantity, that will generate a SPECIFIC volume of carbon dioxide gas UNDER specified conditions of temperature and pressure, here we take one atmosphere and 298∙K .

$nMgCO3={\frac{10.5g}/{84.31∙g∙mol−1}}=0.125∙mol$

And given the equation, should 0.125∙mol magnesium salt be thermally decomposed we get an equivalent molar quantity of magnesium oxide, and CARBON DIOXIDE evolved.

And we know that under standard conditions of temperature and pressure (which you will have to look up), that one mole of idealized gas occupies a 24.5∙L volume.

And so we take the product… $0.125∙mol×24.5∙L∙mol−1≃3∙L.$

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