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 0.610 m nh4i solution is required to react with 229 ml of a 0.640 m pb(no3)2 solution?

Answer : The volume of solution required is 482 mL

Explanation :

The given balanced chemical reaction is:

First we have to calculate the moles of .

Now we have to calculate the moles of

From the balanced chemical reaction we conclude that,

As, 1 mole of react to give 2 moles of

So, 0.147 mole of react to give moles of

Now we have to calculate the volume of

conversion used : (1 L = 1000 mL)

Therefore, the volume of solution required is 482 mL

i'm assuming that the correct answer is indeed the first one/ "a"..

the answer to your question is c