How many milliliters of \(0.383 M \mathrm{HCl}\) are needed to react with \(16.2 \mathrm{~g}\) of \(\mathrm{CaCO}_{3} ?\) $$ 2 \mathrm{HCl}(a q)+\mathrm{CaCO}_{3}(s) \longrightarrow \mathrm{CaCl}_{2}(a q)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) $$

Chemical reactions form the backbone of chemistry. They involve the transformation of reactants into products. In our exercise, \(\text{HCl}\) and \(\text{CaCO}_3\) are the reactants, and the products are \(\text{CaCl}_2\), \(\text{CO}_2\), and \(\text{H}_2\text{O}\). The balanced chemical equation for this reaction is as follows: \[2 \text{HCl}(aq) + \text{CaCO}_3(s) \rightarrow \text{CaCl}_2(aq) + \text{CO}_2(g) + \text{H}_2\text{O}(l)\]. Balancing chemical equations is essential because it ensures the conservation of mass according to the Law of Conservation of Matter. Each side of the equation must have the same number and types of atoms. This specific reaction shows hydrochloric acid reacting with calcium carbonate to produce calcium chloride, carbon dioxide, and water. Understanding the balanced equation allows us to determine the proportions in which these substances react.

The mole concept is fundamental in chemistry for quantifying amounts of a substance. One mole is defined as \(6.022 \times 10^{23}\) entities (Avogadro's number) of any chemical species—atoms, molecules, ions, etc. In our exercise, we first convert the given mass of \(\text{CaCO}_3\) to moles. The molar mass of \(\text{CaCO}_3\) is approximately 100.09 g/mol, so \[\text{Moles of CaCO}_3 = \frac{16.2 \text{ g}}{100.09 \text{ g/mol}} \rightarrow 0.162 \text{ mol}\]. Next, we use the mole ratio from the balanced equation, which shows that 2 moles of \(\text{HCl}\) react with 1 mole of \(\text{CaCO}_3\). Thus, to find the moles of \(\text{HCl}\) required, we calculate \[\text{Moles of HCl} = 2 \times 0.162 \text{ mol} \rightarrow 0.324 \text{ mol}\]. The mole concept helps us understand the stoichiometry of the reaction, ensuring the quantities of reactants and products are accurately determined.

Solution concentration describes how much solute is present in a given quantity of solvent. It's often expressed in terms of molarity (M), which is moles of solute per liter of solution. In our exercise, the concentration of the \(\text{HCl}\) solution is provided as 0.383 M. To find the volume of \(\text{HCl}\) solution needed to react with \(16.2 \text{ g} \text{CaCO}_3\), we rearrange the molarity equation \[ M = \frac{\text{moles}}{\text{liter}}\]. We already calculated the moles of \(\text{HCl}\) needed as 0.324 mol. Plugging in the values, we find: \[ \text{Volume} = \frac{0.324 \text{ mol}}{0.383 \text{ M}} \rightarrow 0.846 \text{ L}\]. Finally, converting liters to milliliters, we get \[0.846 \text{ L} = 846 \text{ mL}\]. Thus, 846 mL of \(0.383 M \text{HCl}\) solution is required. This calculation ensures we use the correct volume of the solution to fully react with the given amount of calcium carbonate.