100 mL of LiNO₃, 0.241 M is mixed with 240 mL of 0.618 M Ca(NO₃)₂. What is the final concentration of NO₃^- in the solution? A) 0.508 M B) 1.01 M C) 0.943 M D) 0.756 M E) 1.38 M

Question

Quizplus · Accepted Answer

Step 1: Calculate moles of LiNO₃ moles = Molarity x volume moles = 0.241 x 0.1 moles = 0.0241 Step 2: Calculate moles of Ca(NO₃)₂ moles = Molarity x volume moles = 0.618 x 0.24 moles = 0.1483 Step 3: Determine limiting reactant LiNO₃ and Ca(NO₃)₂ will react in a 1:1 ratio based on the balanced chemical equation: 2 LiNO₃ + Ca(NO₃)₂ → CaN2 + 4 LiNO3 So, the reactant with the smaller number of moles is the limiting reactant. In this case, LiNO₃ has the smaller number of moles, so it is the limiting reactant. Step 4: Calculate moles of NO₃^- produced From the balanced equation, 2 moles of LiNO₃ produces 4 moles of LiNO3. Therefore, 0.0241 moles of LiNO₃ will produce: moles of NO₃^- = (4/2) x 0.0241 moles of NO₃^- = 0.0482 Step 5: Calculate final concentration of NO₃^- Volume of final solution = 100 mL + 240 mL = 340 mL Molarity of NO₃^- = moles/volume Molarity of NO₃^- = 0.0482/0.34 Molarity of NO₃^- = 0.142 M Therefore, the final concentration of NO₃^- in the solution is 0.142 M, which is closest to option C (0.943 M).

100 ML of LiNO3, 0

100 ML of LiNO₃, 0