AgNO3 + NaI → AgI + NaNO3
From the equation, we can see that 1 mole of silver nitrate reacts with 1 mole of sodium iodide to produce 1 mole of silver iodide and 1 mole of sodium nitrate.
The molar mass of silver nitrate is 169.87 g/mol, and the molar mass of sodium iodide is 149.89 g/mol.
Therefore, 1.7 g of silver is equal to 1.7 g / 169.87 g/mol = 0.01 moles of silver nitrate.
And 1.5 g of sodium iodide is equal to 1.5 g / 149.89 g/mol = 0.01 moles of sodium iodide.
Since the mole ratio between silver nitrate and sodium iodide is 1:1, the limiting reactant in this case is silver nitrate.
Therefore, the maximum amount of sodium nitrate that can be produced is 0.01 moles of sodium nitrate.
The molar mass of sodium nitrate is 84.99 g/mol, so 0.01 moles of sodium nitrate is equal to 0.01 moles * 84.99 g/mol = 0.85 g of sodium nitrate.
Therefore, the amount of sodium nitrate produced when 1.7 g of silver and 1.5 g iodide reacted is 0.85 g.
