91Ó°ÊÓ

Stoichiometry is the study of the quantitative relationships in chemical reactions. It involves calculations based on the balanced chemical equation, ensuring that the same number of each type of atom appears on both sides of the equation. In our exercise, we used stoichiometry to determine the amount of \(\text{HNO}_3\) needed to react completely with \(\text{CuS}\). \[\text {CuS}(s) + 8 \text {HNO}_3 (aq) \rightarrow \text {CuSO}_4 (aq) + 8 \text {NO}_2 (g) + 4 \text {H}_2 \text {O} (l)\] To carry out stoichiometric calculations, we need to use the coefficients from the balanced equation to understand the ratio in which reactants combine and products form. For example, our balanced equation tells us that 1 mole of \(\text{CuS}\) reacts with 8 moles of \(\text{HNO}_3\).

Molar mass is essential in stoichiometry as it relates the mass of a substance to the amount of substance (moles). We calculate the molar mass by adding up the atomic masses of each element in the compound. For \(\text{CuS}\), we have: \(\text{Molar mass of CuS} = 63.55 \text{ g/mol (Cu)} + 32.07 \text{ g/mol (S)} = 95.62 \text {g/mol}\). \[\text{Moles of CuS} = \frac{24.8 \text{ g}}{95.62 \text{ g/mol}} = 0.259 \text{ mol CuS}\] This calculation helps us convert the given mass of \(\text{CuS}\) to moles, which is necessary for further stoichiometric calculations.

Solution concentration, often given in molarity (M), tells you how many moles of solute are present in a liter of solution. The concentration formula is \(\text{M} = \frac{\text{mol of solute}}{\text{L of solution}}\). In our example, the concentration of \(\text{HNO}_3\) is given as 16.0 M. We used this information to find out the volume of \(\text{HNO}_3\) solution needed: \[\text{Volume (L)} = \frac{2.072 \text {mol}}{16.0 \text{ M}} = 0.1295 \text{ L}\]\[\text{Volume (mL)} = 0.1295 \text{ L} \times 1000 \text{ mL/L} = 129.5 \text{ mL}\]

Reactions with acids often produce salts, water, and sometimes gases depending on the reactants involved. In our example, \(\text{HNO}_3\) (a strong acid) reacts with \(\text{CuS}\) forming \(\text{CuSO}_4\) (a salt), \(\text{NO}\) (a gas), and \(\text{H}_2\text{O}\) (water). These reactions involve proton (H\textsuperscript{+}) transfers that cause dissolution of the reactant (CuS) into soluble products: \[\text{CuS} + 8 \text{HNO}_3 \rightarrow \text{CuSO}_4 + 8 \text{NO}_2 + 4 \text {H}_2 \text{O}\]

Copper compounds, such as copper(II) sulfide (CuS) and copper(II) sulfate (CuSO\textsubscript{4}), are common in various chemical reactions. CuS is an insoluble black solid, while CuSO\textsubscript{4} is a blue crystalline solid soluble in water. In our example, CuS reacts with \(\text{HNO}_3\) to form CuSO\textsubscript{4}, proving the transformation of an insoluble compound to a soluble one. This change is vital in many industrial and laboratory processes involving copper: \[\text{CuS} + 8 \text{HNO}_3 \rightarrow \text{CuSO}_4 + 8 \text{NO}_2 + 4 \text{H}_2\text{O}\]