A \( 41-\mathrm{g} \) sample of potassium completely reacts with chlorine to form 78 g of potassium chloride. How many grams of chlorin must have reacted? \( \begin{array}{l}37 \mathrm{~g} \\ 41 \mathrm{~g} \\ 17 \mathrm{~g} \\ 78 \mathrm{~g}\end{array} \)

To determine how many grams of chlorine reacted with the potassium, we can use the law of conservation of mass, which states that the mass of the reactants must equal the mass of the products. We know: - Mass of potassium (K) = 41 g - Mass of potassium chloride (KCl) produced = 78 g Let \( m \) be the mass of chlorine (Cl) that reacted. According to the conservation of mass: \[ \text{Mass of reactants} = \text{Mass of products} \] This can be expressed as: \[ \text{Mass of potassium} + \text{Mass of chlorine} = \text{Mass of potassium chloride} \] Substituting the known values: \[ 41 \, \text{g} + m = 78 \, \text{g} \] Now, we can solve for \( m \): \[ m = 78 \, \text{g} - 41 \, \text{g} = 37 \, \text{g} \] Thus, the mass of chlorine that reacted is \( \boxed{37 \, \text{g}} \).