Question

You carefully weigh out 18. 00 g of caco3 powder and add it to 72. 90 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 83. 52 g. The relevant equation is.

Answer 1

The relevant equation is $CaCO_{3} (s)+2HCl(aq)=H_{2}O (l)+CO_{2} (g)+CaCl_{2} (aq)$, assuming no other reactions take place during this reaction, with a mass of carbon dioxide of 8.49 g.

$CaCO_{3} (s)+2HCl(aq)=H_{2}O (l)+CO_{2} (g)+CaCl_{2} (aq)$

1 mole        2 moles 1mole    1 mole        1 mole

The moles are from the reaction, find the moles from the reaction to determine the limiting agent mass of CaCO3 = 17.00g

Number of moles of CaCO3 = (17.00g) (88.075g/mole) = 0.193 mole)

Number of moles of HCl = (68.85g) (36.45g/mole) = 1.889 mole

From calculations, (moles of HCl) (moles of CaCO3) = 1.889/0.193 = 10/1

From the reaction, the number of 1 mole of CaCO3 reacted to produce 1 mole of CO2

From calculations, the number of 0.193 moles of CaCO3 must produce 0.193 moles of CO2

Mass of CO2 = (molecular massCO2)(moles of CO2)

 = 8.49 g

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Although part of your question is missing, you might be referring to this full question:

You carefully weigh out 17.00 g of CaCO3 powder and add it to 68.85 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 78.88 g. The relevant equation is

CaCO3(s)+2HCl(aq)→H2O(l)+CO2(g)+CaCl2(aq)

Assuming no other reactions take place, what mass of CO2 was produced in this reaction?