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Report (any type)/Brief report
22/09/2021
##”645 mg”##
Your starting point here will be the balanced chemical equation for this .
Calcium carbonate, ##”CaCO”_3##, will react with hydrochloric acid, ##”HCl”##, to produce aqueous calcium chloride, ##”CaCl”_2##, water, and carbon dioxide, ##”CO”_2##
##”CaCO”_ (3(s)) + color(red)(2)”HCl”_ ((aq)) -> “CaCl”_ (2(aq)) + “H”_ 2″O”_ ((l)) + “CO”_ (2(g)) uarr##
Notice that it takes ##color(red)(2)## moles of hydrochloric acid to neutralize ##1## mole of calcium carbonate. This means that the reaction will always consume twice as many moles of hydrochloric acid than of calcium carbonate.
The problem provides you with the and volume of the hydrochloric acid solution. This means that you can use the definition of molarity to find how many moles of acid it contained
##color(blue)(c = n/V implies n = c * V)##
##n_(HCl) = “0.547 M” * 24.65 * 10^(-3)”L” = “0.013484 moles HCl”##
Use the aforementioned mole ratio to find the number of moles of calcium carbonate that must have been present in the Rolaids table
##0.013484 color(red)(cancel(color(black)(“moles HCl”))) * (“1 mole CaCO”_3)/(color(red)(2)color(red)(cancel(color(black)(“moles HCl”)))) = “0.0067420 moles CaCO”_3##
Finally, to find the mass of calcium carbonate that contains this many moles, use the compound’s molar mass
##0.0067420 color(red)(cancel(color(black)(“moles CaCO”_3))) * “100.1 g”/(1color(red)(cancel(color(black)(“mole CaCO”_3)))) = “0.645 g”##
Expressed in milligrams, the answer will be
##0.645 color(red)(cancel(color(black)(“g”))) * (10^3″mg”)/(1color(red)(cancel(color(black)(“g”)))) = color(green)(“645 mg”)##
The answer is rounded to three .
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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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