Lead(ii) nitrate and ammonium iodide react to form lead(ii) iodide and ammonium nitrate according to the reaction pb(no3)2(aq)+2nh4i(aq)⟶pbi2(s)+2nh4 no3(aq) what volume of a 0.290 m nh4i solution is required to react with 253 ml of a 0.680 m pb(no3)2 solution?

The balanced chemical reaction between and is as follows:

From the above reaction, it is clear that 1 mole of reacts with 2 moles of .

Molarity and volume of is 0.680 M and 253 mL respectively.

Calculate number of moles of as follows:

Here, n is number of moles and V is volume. Convert volume from mL to L

Thus,

Rearranging equation and putting values,

Since, 1 mol of reacts with 2 moles of thus, 0.172 mol of will react with of .

Now, molarity of is 0.290 M or mol/L thus, volume can be calculated as:

Converting volume into mL

Therefore, volume of 0.290 M  required will be 1186 mL.

- lead + phosphoric acid = hydrogen + lead(ii) phosphate