Percent composition is determined by dividing the mass of the specified compound by the total mass, then multiply by 100. $"Percent composition"="mass of specified component"/"total mass"xx100"$ Add the masses of...
1 Answers 1 viewsThe balanced chemical equation The molar mass of $"CaCO"_3$ The molar ratio of $"CaO"$ to $"CaCO"_3$ The molar mass of $"CaO"$ Your strategy has four steps:...
1 Answers 1 viewsWe need a stoichiometric reaction that represents the decomposition of calcium carbonate: $CaCO_3(s) + Delta rarr CaO(s) + CO_2(g)uarr$ And thus calcium oxide and calcium carbonate are present in equimolar...
1 Answers 1 views$6.022xx10^23$ units of ANYTHING, specifies 1 mole of that thing, i.e. $N_A$ individual units of that thing........... And thus in $1*mol$ of $CaCO_3$, $N_A$ formula units, there are: $N_A$...
1 Answers 1 viewsYou take the atomic mass of calcium over the atomic mass of $CaCO_3$ and multiply it by 100 to get the percentage. The mass of $Ca$ is 40.078 g; the...
1 Answers 1 viewsSo we have the $"%calcium carbonate, "m/m$. We are not finished yet, because we have to address the percentage by mass of calcium metal in calcium carbonate: $%Ca=(40.08*g*mol^-1)/(100.09*g*mol^-1)=40%$ with respect...
1 Answers 1 viewsThe equation says $CaCO_3 + 2HCl$, which means that for every $2$ moles of $HCl$ there is one of $CaCO_3$. This is called the molar ratio, in this case...
1 Answers 1 viewsWe interrogate the stoichiometric equation...... $CaCO_3(s) + Delta rarr CaO(s) + CO_2(g) uarr$ We require a molar quantity of $(15.0*g)/(56.08*g*mol^-1)=0.268*mol,$ with respect to $"calcium oxide"$. And clearly, we need a...
1 Answers 1 views$CaCO_3(s) + Delta rarr CaO(s) + CO_2(g)uarr$ There is thus a 1:1 molar equivalence of carbonate to carbon dioxide. $"Moles of calcium carbonate"$ $=$ $(235.0*g)/(100.0869*g*mol^-1)$ $=$ $2.35$ $mol$. So, by...
1 Answers 1 views$10gramsCaCO_3(s) => CaO(s) + (?)O_2(g)$ $"moles" CaCO_3 = ((10g)/(100(g/(mol))))$ = $0.10"mole"$ From reaction ratios => moles $CaCO_3(s)$ consumed = moles $O_2(g)$ produced = 0.10 mole $(O_2(g))$**** If a 'dry'...
1 Answers 1 views