Problemas
25. Consider the reaction of siver metal and nitric acid (HNO3) given in the equation below What volume of a 1.85 Mnitric acid solution is needed to completely react with silver? Balanced Equation 3Ag+4HNO3arrow 3AgNO3+NO+2H2O 1.1161 6.0051 0,006 L 51601 Clear my selection Multiple Chore 1 point
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Christian
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To solve this problem, we need to determine the volume of the 1.85 M nitric acid solution required to completely react with 0.334 grams of silver.Given information:- Balanced equation: 3Ag + 4HNO3 → 3AgNO3 + NO + 2H2O- Mass of silver: 0.334 gramsStep 1: Calculate the number of moles of silver.Moles of silver = Mass of silver / Molar mass of silverMoles of silver = 0.334 g / 107.87 g/mol = 0.00309 molStep 2: Determine the number of moles of nitric acid required.According to the balanced equation, 3 moles of silver react with 4 moles of nitric acid.Moles of nitric acid required = (4/3) × Moles of silverMoles of nitric acid required = (4/3) × 0.00309 mol = 0.00412 molStep 3: Calculate the volume of the 1.85 M nitric acid solution.Volume of nitric acid solution = Moles of nitric acid / Molarity of nitric acid solutionVolume of nitric acid solution = 0.00412 mol / 1.85 M = 0.0061 LTherefore, the correct answer is 0.0061 L.