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) + 2nh4no3 (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?

Question

Marin Cunningham

Chemistry

23 February, 02:18

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) + 2nh4no3 (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?

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Raquel Garrison · Answer 1

0.229 L * 0.640 mol Pb (No3) 2/L = 0.147 mol Pb (No3) 2

by equation 1 mol Pb (No3) 2 requires 2 mol NH4I

so 0.147 mol Pb (No3) 2 requires 2*0.147 mol NH4I

2*0.147 mol NH4I = 0.294 mol NH4I

0.294 mol NH4I * 1L/0.610 mol = 0.482 L = 482 ml of the 0.610 m NH4i solution