Question #54ef3

1 Answer
Feb 1, 2018

Determine moles of AgI formed, that is same as number of moles of AgNO3, then divide moles by 0.050 L

Explanation:

AgNO3(aq) + HI(g) --> AgI (s) + HNO3(aq)

1 mole of silver nitrate yields 1 mole of silver iodide
molar mass of Ag I = 234.77g/mol

mole AgI #= (2.35g)( (1mol)/(234.77g))#

#mol AgI = 0.0100 mol#

there is one mole of Ag in each mole of AgI so there was 0.0100 mol of silver in the original 50.0 mL solution

Molarity is mol/L

Molarity #= (0.0100mol)/(0.050L)#

Molarity #= 0.20 M#