Lead(ii) nitrate and ammonium iodide react to form lead(ii) iodide and ammonium nitrate according to the reaction. what volume of a 0.530 m nh4i solution is required to react with 339 ml of a 0.460 m pb(no3)2 solution?

The equation that describes this reaction is:
Pb(NO3)2 + 2 NH4I → PbI2 + 2 NH4NO3

molarity = number of moles / volume
We will use this equation to get the number of moles of Pb(NO3)3 as follows:
0.46 = number of moles / (339 * 10-3)
number of moles = 0.15594 moles

From the balanced equation above, we can note that 1 mole of Pb(NO3)2 reacts with 2 moles of NH4I. Therefore, we can simply use cross multiplication to find the number of moles of NH4I that reacts with 0.15594 moles of Pb(NO3)2 as follows:
number of moles of NH4I = 0.15594 * (2/1) = 0.31188 moles

Finally, we will calculate the volume of the required NH4I as follows:
molarity = number of moles / volume
volume = number of moles / molarity
volume of NH4I = 0.31188 / 0.53 = 0.58845 liters

Sorry i dont understand the question can you rephase it?