Ow many grams of calcium chloride will be produced when 29.0 g of calcium carbonate are combined with 10.0 g of hydrochloric acid?

15.2 g Begin by calculating the molar mass of the reactants and the desired product. First, lookup the atomic weights of each element involved. Calcium = 40.078 Carbon = 12.0107 Oxygen = 15.999 Hydrogen = 1.00794 Chlorine = 35.453 Now sum up the products of the atomic weights of each element times the number of times each element is used in each compound Calcium Chloride (CaCl2) 40.078 + 2 * 35.453 = 110.984 Calcium Carbonate (CaCO3) 40.078 + 12.0107 + 3 * 15.999 = 100.0857 Hydrochloric acid (HCl) 1.00794 + 35.453 = 36.46094 Figure out how many moles of each reactant is available by dividing mass by molar mass Calcium Carbonate (CaCO3) 29.0 g / 100.0857 g/mol = 0.289752 mol Hydrochloric acid (HCl) 10.0 g / 36.46094 g/mol = 0.274266 mol Create a balanced equation for the reaction CaCO3 + 2 HCl => CaCl2 + CO2 + H2O Looking at the balanced equation, it takes 2 moles of HCl for each mole of CaCO3, since the number of available moles for each reactant is about equal, the limiting reactant will be HCl. So for each mole of HCl, 0.5 moles of CaCO3 will be needed. 0.5 * 0.274266 = 0.137133 So we'll be producing 0.137133 moles of CaCl2. Just multiply by the previously calculated molar mass 0.137133 mol * 110.984 g/mol = 15.21956887 g Since we only have 3 significant digits available in our measurements, round the result to 3 significant digits. 15.21956887 g = 15.2 g

Neither model is able to show the reactivity of potassium.