Final answer:
To find the mass of silver reacting with nitric acid, we use stoichiometry to calculate moles from the given concentration and volume of nitric acid, then apply the mole ratio from the balanced equation, and finally convert moles of silver to grams.
Step-by-step explanation:
The question involves calculating the mass of silver that reacts with a given volume and concentration of nitric acid in a redox reaction. To find the answer, we follow a series of stoichiometric steps:
- First, we need to clarify the balanced chemical equation for the reaction between silver (Ag) and nitric acid (HNO3). However, the provided chemical equation in the question appears to be missing some coefficients and possibly has a typo. Assuming the corrected equation is 3 Ag(s) + 4 HNO3(aq) → 3 AgNO3(aq) + NO(g) + 2 H2O(l), this would involve silver reacting with nitric acid to produce silver nitrate, nitrogen monoxide, and water.
- We then use the molarity (M) of nitric acid and the volume (mL) provided to calculate the number of moles of HNO3 used in the reaction using the formula moles = M × Volume (in liters).
- Next, we utilize the mole ratio from the balanced equation to determine the moles of silver that would react with the calculated moles of HNO3.
- Finally, we convert the moles of Ag to grams using the formula mass = moles × molar mass of silver.
In this scenario, with 42.50 mL of 12.0M nitric acid, first convert milliliters to liters: 42.50 mL = 0.04250 L. Calculating the moles of HNO3: 0.04250 L × 12.0 mol/L = 0.510 mol. According to the balanced equation (assuming it's corrected), the mole ratio of Ag to HNO3 is 3:4. So, the moles of Ag reacting would be (3/4) × 0.510 mol = 0.3825 mol. Finally, with the molar mass of Ag being approximately 107.87 g/mol, the mass of Ag reacting would be 0.3825 mol × 107.87 g/mol = 41.25 grams of silver.