CaI2+2AgNO3⟶2AgI+Ca(NO3)2 CaI 2 + 2 ⁢ AgNO 3 ⟶ 2 AgI + Ca ( NO 3 ) 2 what mass of silver iodide is produced from 5.31 mol of calcium iodide?

Answer 1

CaI₂(aq) + 2AgNO₃(aq) ==> 2AgI(s) + Ca(NO₃)₂(aq) ... balanced equation

molar mass AgI = 234.8 g/mol

5.31 mol CaI₂ x 2 mol AgI / 1 mol CaI₂ x 234.8 g AgI / mol = 2494 g AgI