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

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.<br /><br />Given information:<br />- Balanced equation: 3Ag + 4HNO3 → 3AgNO3 + NO + 2H2O<br />- Mass of silver: 0.334 grams<br /><br />Step 1: Calculate the number of moles of silver.<br />Moles of silver = Mass of silver / Molar mass of silver<br />Moles of silver = 0.334 g / 107.87 g/mol = 0.00309 mol<br /><br />Step 2: Determine the number of moles of nitric acid required.<br />According to the balanced equation, 3 moles of silver react with 4 moles of nitric acid.<br />Moles of nitric acid required = (4/3) × Moles of silver<br />Moles of nitric acid required = (4/3) × 0.00309 mol = 0.00412 mol<br /><br />Step 3: Calculate the volume of the 1.85 M nitric acid solution.<br />Volume of nitric acid solution = Moles of nitric acid / Molarity of nitric acid solution<br />Volume of nitric acid solution = 0.00412 mol / 1.85 M = 0.0061 L<br /><br />Therefore, the correct answer is 0.0061 L.