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)#?

Answer 1

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

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?

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Answer 2

The balanced chemical equation for the reaction is:

[ \text{CaCO}_3 (s) + 2 \text{HCl} (aq) \rightarrow \text{CaCl}_2 (aq) + \text{CO}_2 (g) + \text{H}_2\text{O} (l) ]

Using the molar mass of CaCO3 (100.09 g/mol), the moles of CaCO3 can be calculated as ( \frac{20 \text{ g}}{100.09 \text{ g/mol}} ).

According to the balanced equation, 1 mole of CaCO3 produces 1 mole of CO2. Therefore, the moles of CO2 produced will be the same as the moles of CaCO3.

The volume of CO2 at standard temperature and pressure (STP) can be calculated using the ideal gas law: ( V = n \times 22.4 \text{ L/mol} ), where ( n ) is the moles of CO2.

Finally, the volume can be converted from liters to milliliters ((1 \text{ L} = 1000 \text{ mL})).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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