Question

You carefully weigh out 13.00 g of CaCO3 powder and add it to 52.65 g of HCl solution. You notice bubbles as a reaction takes place. You then weigh the resulting solution and find that it has a mass of 60.32
g. The relevant equation isCaCO3(s)+2HCl(aq)?H2O(l)+CO2(
g.+CaCl2(aq)Assuming no other reactions take place, what mass of CO2 was produced in this reaction?Express your answer to three significant figures and include the appropriate units.

Answer 1

CaCO₃ + 2HCl = CaCl₂ + CO₂ + H₂O

n(CaCO₃)=m(CaCO₃)/M(CaCO₃)
n(CaCO₃)=13.00/100.09=0.1299 mol

Δm=13.00+52.65-60.32=5.33 g

m(CO₂)=5.33 g
n(CO₂)=5.33/44.01=0,1211 mol

w=0.1211/0.1299=0,9323 (93.23%)

Answer 2

Answer:

The amount of carbon dioxide produced is 5.33 gr.

Explanation:

The easiest way to solve this problem it by knowing that the mass of a sistem is always the same (the mass can't be created or destroyed, only converted).

Following that statement, we can substract from the final mass of the reaction (60.32gr), the mass of the reactives (13 gr and 52.65 gr), and keeping in mind that the only mass that can't be weight is the mass of CO₂ because it leaves the sistem in the form of a gas:

Δm = 60.32-13.00-52.65=-5.33 g

The negative indicates a lack of mass, in this case the CO₂  wich has been produced.

If we look at the reaction equation, we can see that for each mol of CaCO₃, it is produced one mol of CO₂.

CaCO₃ + 2HCl = CaCl₂ + CO₂ + H₂O

Knowing the amount of CaCO₃ (13 gr) and  searching in the periodic table to  find its molar mass, we can say that we have 0.12988 mol of CaCO₃.

Doing the same steps for carbon dioxide, we can see that only 0.1211  mol of CO₂ have been produced. That indicates that te reaction has not been completed.