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Answer 1 · 2018-02-01T20:00:20

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

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 $= (2.35 g) (\frac{1 m o l}{234.77 g})$ $= (2.35g)( (1mol)/(234.77g))$

$m o l A g I = 0.0100 m o l$ $mol AgI = 0.0100 mol$

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 $= \frac{0.0100 m o l}{0.050 L}$ $= (0.0100mol)/(0.050L)$

Molarity $= 0.20 M$ $= 0.20 M$