Given a #20*g# mass of calcium carbonate, how much carbon dioxide will be generated, when it is added to a #100*mL# volume of #"20% w/w"# solution of #HCl(aq)#?

1 Answer
Aug 19, 2017

Answer:

The units of concentration of the hydrochloric acid are unfortunate.....I gets approx. a #5*L# volume under standard conditions.

Explanation:

We interrogate the reaction......

#CaCO_3(s) + 2HCl(aq) rarr CaCl_2(aq) + H_2O(l) + CO_2(g)uarr#

Now this site reports that the density of #20%# #HCl# is #rho=1.098*g*mL^-1#. These data SHOULD have been supplied with the question.

And thus moles of #HCl=(20%xx100*cancel(mL)xx1.098*cancelg*cancel(mL^-1))/(36.46*cancelg*mol^-1)#

#=0.602*1/(mol^-1)=0.602*1/(1/(mol))=0.602*mol#

And thus the molar concentration of the hydrochloric acid was approx. #6*mol*L^-1#.

And also moles of #"calcium carbonate"-=(20*g)/(100.09*g*mol^-1)#

#=0.200*mol#.

There is thus a stoichiometric excess of hydrochloric acid, and calcium carbonate is the limiting reagent. A #0.200*mol# quantity of carbon dioxide gas will be released.

Depending on your syllabus, we know that the molar volume at #1*atm# and #298*K# is #24.5*L*mol^-1#. You have to check with the given definitions in your course.

And thus #"volume"=24.5*L*mol^-1xx0.200*mol=??*L.#

How much acid is in excess?