Print Print Calculator Periodic Table Question 5 of 72 Map General Chemistr Univ

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Print Print Calculator Periodic Table Question 5 of 72 Map General Chemistr University Science Books presented by Sapling Lea Donald McQuarrie. Peter A Rock .Ethan Gallogly When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced Caco, (s) HCI (aq) Cacia (aq) H,0(I) co How 15.0 g grams of calcium chloride will be produced when 25.0 g of calcium carbonate are combined with many of hydrochloric acid? Number g CaCI, Which reactant is in excess and how many grams of this reactant will remain after the reaction is complete? Number O CaCO3. g of O HCI o Previous Give Up & View Solution Check Answer Next Ext Hint

Explanation / Answer

Solution.

The molar mass of CaCO3 is 100.0869 g/mol; its amount is n = m/M = 25.0/100.0869=0.25 moles;

The molar mass of HCl is 36.46094 g/mol; its amount is n = m/M = 15.0/36.46094=0.411 moles;

As 0.25 moles of  CaCO3 require 0.25*2 = 0.50 moles of HCl (but we have 0.411), HCl is the limiting reactant. Therefore, CaCO3 will remain after the reaction complete.

The amount of CaCl2 formed is 0.411/2 = 0.2055 moles. The molar mass of CaCl2 is 110.984, therefore, the mass is 110.984*0.2055 = 22.8 g.

CaCO3 is in excess. The amount of reacted CaCO3 is 0.2055 moles, or 0.2055*100.09 = 20.568 g.

The remaining mass is 25.0-20.6 = 4.40 g of CaCO3 .

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