What mass of silver iodide can be made by the reaction of 10.0 g of silver nitrate (AgNO3) with 10.0 g of sodium iodide (NaI)?Ag+(aq)  +  I-(aq)  ®  AgI(s)

To solve this problem, we need to use stoichiometry, which is a method in chemistry that uses relationships between reactants and/or products in a chemical reaction to determine desired quantitative data.

Here are the steps:

1. First, we need to find the molar mass of both silver nitrate (AgNO3) and sodium iodide (NaI). The molar mass of AgNO3 is approximately 169.87 g/mol and the molar mass of NaI is approximately 149.89 g/mol.

2. Next, we convert the mass of the reactants to moles. We do this by dividing the given mass by the molar mass.

   For AgNO3: 10.0 g / 169.87 g/mol = 0.0588 mol For NaI: 10.0 g / 149.89 g/mol = 0.0667 mol

3. The balanced chemical equation tells us that the ratio of AgNO3 to NaI to AgI is 1:1:1. This means that the reaction will run out of AgNO3 first because we have less of it. Therefore, AgNO3 is the limiting reactant.

4. Since the ratio of AgNO3 to AgI is 1:1, the moles of AgI produced is equal to the moles of the limiting reactant (AgNO3). So, we have 0.0588 mol of AgI.

5. Finally, we convert the moles of AgI to grams. The molar mass of AgI is approximately 234.77 g/mol.

   0.0588 mol * 234.77 g/mol = 13.8 g

So, 13.8 g of silver iodide can be made by the reaction of 10.0 g of silver nitrate with 10.0 g of sodium iodide.