4-6. When calcium carbonate is added to hydrochloric acid, calcium chloride, car

Question

4-6. When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced according to the balanced chemical equation below: CaCO3(s) + 2HC1(aq) 4 CaC12(aq) +H20(1) + CO2(g) Imagine mixing 2.5 moles of calcium carbonate with 4.8 moles of hydrochloric acid. Answer the questions about the chemical reaction that occurs. 4. What species is the limiting reagent? a. Calcium carbonate b. Hydrochloric acid c. Calcium chloride d. Water e. Carbon dioxide 5. What are the maximum number of moles of carbon dioxide that can be produced? a. 2.4 moles b. 2.5 moles c. 4.8 moles d. 5.0 moles 6. Which equation below represents the calculation for the number of grams of calcium chloride are produced if the reaction proceeds with a 65% yield? a. 2.4 moles CaCl2 * 110.98 grams/mol * 0.65 b. 2.5 moles CaCl2* 110.98 grams/mol * 0.65 c. 2.4 moles CaCl2* 110.98 grams/mol / 0.65 d. 2.5 moles CaCl2* 110.98 grams/mol / 0.65

Explanation / Answer

4)

Solve for the moles of one the products from the moles of both reactants (HCl and CaCO3) to figure out the limiting reagent. I'll solve for the moles of CO2.

Moles of CO2 = 2.5 moles of CaCO3 x 1 mole of CO2/1 mole of CaCO3 = 2.5 moles

Moles of CO2 = 4.8 moles of HCl x 1 mole of CO2/2 moles of HCl = 2.4 moles (theoretical moles)

Since HCl produces the fewer moles of CO2, it is the limiting reagent.

5)

I solved this in question 4. The maximum moles is the theoretical moles which is 2.4.

6)

Theoretical moles of CaCl2 = 2.4 moles (since ratio of HCl:CaCl2 is 2:1 just like the ratio above for HCl:CO2).

Theoretical mass of CaCl2 = (2.4 moles x 110.98 g/mol)

% yield = (actual yield / theoretical yield) x 100%

0.65 = [actual yield / (2.4 moles x 100.98 g/mol)] ... cross multiply to solve for the actual yield

actual yield = (0.65) (2.4 moles x 100.98 g/mol)

Hope that helps! :)

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