CaCO3 + 2HCl ==> CaCl2 + H2O + CO2
Convert 29.0g CaCO3 to mols. mol = grams/molar mass.
Convert 13.0g HCl to mols.
Using the coefficients in the balanced equation, convert mols CaCO3 to mols CaCl2.
Do the same for mols HCl to mols CaCl2.
It is likely that the two values will not agree which means one of them is wrong; the correct value in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent. The other reagent is the one in excess.
g CaCl2 produced is the smaller value for mols x molar mass CaCl2 = ?
Now that you know the excess and limiting reagents, use the coefficients to convert mols of the limiting reagent used to mols excess reagent used, subtract mols excess initially - mols excess reagent used = mols excess reagent remaining. The convert that to grams by g = mols x molar mass.
When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced.
CaCO3 +2HCl yields CaCl2 +H2O +CO2
How many grams of calcium chloride will be produced when 29.0 g of calcium carbonate are combined with 13.0 g of hydrochloric acid?
Which reactant is in excess and how many grams of this reactant will remain after the reaction is complete?
1 answer