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. (contnd at description)

Question

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?
Express your answer to three significant figures and include the appropriate units.

 

 

My solution

(17+68.85)-78.88=6.97 is not the answer! help please!

Answer 1

Hi Pia,

 

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

1 mole        2 mole          1mole    1mole        1 mole

These moles are from reaction, find the moles from the reaction to determine the limiting agent 

mass of CaCO₃ = 17.00g

molecular mass of CaCO₃ =40.078+3*15.999=88.075 g/mole

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

molecular mass of CaCO3 =40.078+3*15.999=88.075 g/mole

mass of HCl = 68.85 g

molecular mas of HCl = 1.008 +35.45 = 36.45g /mole
Number of moles of HCl = (68.85g)(36.45g/mole) = 1.889 mole 

From reaction, (moles of HCl)(moles of CaCO₃) = 2/1

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

This amount is 5 times more than can react with CaCO₃, so CaCO₃ is the limiting reagent

molecular mass of CO₂ = 12.011 + 2*15.999 = 44.009 g/mole CO₂

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

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

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

                  = (44.009 g/mole)(0.193 mole).

                  = 8.49 g

Let me know if there is error.