A 2.3995 g mixture containing malachite and an inertingredient lost 0.2997 g of

Question

A 2.3995 g mixture containing malachite and an inertingredient lost 0.2997 g of H2O and CO2 whendecomposed by heating to constant mass. A) how many moles of malachite in the mixture produced thismass of CO2 and H2O? B) How many grams of malachite were present in themixture? This is the equation:CuCO3*Cu(OH)2(s)--> 2 CuO (s)+H2O (g) +CO2 A 2.3995 g mixture containing malachite and an inertingredient lost 0.2997 g of H2O and CO2 whendecomposed by heating to constant mass. A) how many moles of malachite in the mixture produced thismass of CO2 and H2O? B) How many grams of malachite were present in themixture? This is the equation:CuCO3*Cu(OH)2(s)--> 2 CuO (s)+H2O (g) +CO2

Explanation / Answer

A) CuCO3.Cu(OH)2 (s) -> 2 CuO (s) +H2O (g) + CO2 (g)         x                                                   x             x Let's call x the number of moles of malachite, you can see that xmol of malachite produces x mol of H2O and x mol ofCO2. Mass of H2O = moles x molar mass = 18.00x (grams) Mass of CO2 = moles x molar mass = 44.01x (grams) Total mass of H2O and CO2 = 18.00x+ 44.01x = 0.2997 (grams) => 62.01x = 0.2997 => x = 0.004833 (mol) = 4.833 x10-3 mol So there are 4.833 x 10-3 moles of malachite in themixture. B) We already have numbers of mole of malachite, we now just needto convert it to grams: Mass of malachite = moles x molar mass = 0.004833 mol x 221.09g/mol = 1.069 g

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