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
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
Molarity