Consider the following reaction equation: CaCO\(_{3(s)}\) + 2HCI\(_{(aq)}\) → CaCl\(_{2(aq)}\) + H\(_2\)O\(_{(l)}\) + CO\(_{2(g)}\). What mass of CaCI\(_2\)...

To solve this problem, we need to use stoichiometry to relate the amount of CaCO\(_3\) reacted to the amount of CaCl\(_2\) produced. First, we need to balance the equation to make sure that the number of atoms of each element is equal on both sides: CaCO\(_3\) + 2HCl → CaCl\(_2\) + H\(_2\)O + CO\(_2\) Next, we need to determine the mole ratio between CaCO\(_3\) and CaCl\(_2\) from the balanced equation: 1 mole CaCO\(_3\) : 1 mole CaCl\(_2\) Now, we can use this mole ratio to calculate the number of moles of CaCl\(_2\) produced from 25.0g of CaCO\(_3\): 25.0 g CaCO\(_3\) x (1 mole CaCO\(_3\)/100 g CaCO\(_3\)) x (1 mole CaCl\(_2\)/1 mole CaCO\(_3\)) = 0.250 mole CaCl\(_2\) Finally, we can use the molar mass of CaCl\(_2\) to calculate the mass produced: 0.250 mole CaCl\(_2\) x 111 g CaCl\(_2\)/1 mole CaCl\(_2\) = 27.8 g CaCl\(_2\) Therefore, the mass of CaCl\(_2\) produced when 25.0g of CaCO\(_3\) is reacted with excess HCl\(_{(aq)}\) is 27.8 g. The correct answer is.