Answer to Question #97473 in Chemistry for Kristin

Question

Calculate the maximum mass of NaIO3 that can be dissolved in 3.00L of 0.200M AgNO3 solution without forming a precipitate

Answer 1

Answer on Question #97473 – Chemistry – Other

Task:

Calculate the maximum mass of NaIO₃ that can be dissolved in 3.00L of 0.200M AgNO₃ solution without forming a precipitate.

Solution:

NaIO₃ + AgNO₃ = AgIO₃ + NaNO₃

1) n(NaIO₃) = n(AgNO₃);

n(NaIO₃) = C(AgNO₃)*V(AgNO₃) = 0.200M * 3.00L = 0.600 mol;

2) n(NaIO₃) = m(NaIO₃) / M(NaIO₃);

M(NaIO₃) = Ar(Na) + Ar(I) + 3*Ar(O) = 22.9898 + 126.9045 + 3*15.999 = 197.8913 (g/mol).

3) m(NaIO₃) = n(NaIO₃)*M(NaIO₃) = 0.600 mol * 197.8913 g/mol = 118.7348 g.

m_max(NaIO₃) = 118.7348 g.

Answer: m_max(NaIO₃) = 118.7348 g.