I want to thank your team (Jeff) for finding a new writer when you were unable to reach my existing writer to revise my paper. I appreciate your professionalism and efforts. Thanks
Report (any type)/Brief report
22/09/2021
I got ##3.93xx10^3 “kg CaCO”_3##.
You didn’t write a reaction. But you probably meant this:
##”CaCO”_3(s) stackrel(Delta)(->) “CaO”(s) + “CO”_2(g)##
where ##Delta## indicates high heat, and ##”CaCO”_3## (calcium carbonate) has decomposed into ##”CaO”(s)## (calcium oxide) and ##”CO”_2(g)## (gaseous carbon dioxide).
You can start by assuming ##”CO”_2(g)## is an ideal gas. At STP, we are at a temperature of ##0^@ “C”##, or ##”273.15 K”##, and ##”1 atm”## of pressure. We know the volume produced is ##”881 L”## of gas.
(The volume of solid is negligibly small in comparison.)
We are not told anything else, so we have to determine the volume in ##bb”1 mol”## (the molar volume) of ideal gas to figure out how many ##”mol”##s of ##”CO”_2## we made.
Recall the following equation for ideal gases:
##bb(PV = nRT)## (ideal gas law)
where ##P##, ##V##, ##n##, ##R##, and ##T## are
respectively.
Rearrange to get the molar volume, ##V/n##:
##V/n = (RT)/P##
##= ((“0.082057 L”cdot”atm/mol”cdot”K”)(“273.15 K”))/(“1 atm”)##
##~~## ##”22.414 L”##.
So, the ##”mol”##s of ##”CO”_2## we have is:
##881 cancel”L” xx “1 mol”/(22.414 cancel”L”) ~~ “39.31 mols CO”_2##
So, back-calculations will give the ##bb”mol”##s of ##”CaCO”_3## needed, and therefore the mass using its molar mass.
##39.31 cancel(“mols CO”_2) xx (“1 mol CaCO”_3(s))/(cancel(“1 mol CO”_2(g)))##
##~~ “39.31 mols CaCO”_3(g)##
So, we have to have used this much mass:
##color(blue)(m_(“CaCO”_3(s))) = 39.31 cancel(“mols CaCO”_3(s)) xx “100.088 g”/(cancel(“1 mol CaCO”_3(s)))##
##~~## ##”3934.1 g”##
##~~## ##color(blue)(3.93xx10^3 “kg CaCO”_3)##
where molar mass is just the sum of the atomic masses for each atom in the compound.
I want to thank your team (Jeff) for finding a new writer when you were unable to reach my existing writer to revise my paper. I appreciate your professionalism and efforts. Thanks
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