When 43.0mL of 5.00gL AgNO3 is added to a coffee-cup calorimeter containing 43.0mL of 5.00gL NaI, with both solutions at 25°C, what mass of AgI forms? Be sure your answer has the correct number of significant figures.gAgI

To find the mass of AgI that forms, we first need to determine the limiting reactant. The reaction is:

AgNO3 + NaI → AgI + NaNO3

From the problem, we know that we have 43.0 mL of 5.00 g/L AgNO3 and 43.0 mL of 5.00 g/L NaI.

First, we need to convert the volumes and concentrations to masses:

Mass of AgNO3 = Volume x Concentration = 43.0 mL x 5.00 g/L = 0.215 g Mass of NaI = Volume x Concentration = 43.0 mL x 5.00 g/L = 0.215 g

Next, we need to convert the masses to moles using the molar masses of AgNO3 and NaI:

Moles of AgNO3 = Mass / Molar mass = 0.215 g / 169.87 g/mol = 0.00126 mol Moles of NaI = Mass / Molar mass = 0.215 g / 149.89 g/mol = 0.00143 mol

From the balanced chemical equation, we know that 1 mole of AgNO3 reacts with 1 mole of NaI to form 1 mole of AgI. Therefore, AgNO3 is the limiting reactant because we have fewer moles of it.

Finally, we can find the mass of AgI that forms by multiplying the moles of the limiting reactant by the molar mass of AgI:

Mass of AgI = Moles x Molar mass = 0.00126 mol x 234.77 g/mol = 0.296 g

So, the mass of AgI that forms is 0.296 g.