Answer to Question #151654 in General Chemistry for Liam Knutson

Question

1. 2.50 L of 0.450 mol/L solution of silver nitrate is mixed with a solution prepared by dissolving 86.2 g sodium phosphate.
A) Which reactant is the limiting Reagent? Show work
B) what chemical amount (moles) of the excess will remain?
C) what mass of precipitate will be produced?

Unit - Stoichiometry

Answer 1

3AgNO₃ + Na₃PO₄ → Ag₃PO₄ + 3NaNO₃

A) n(AgNO₃) $= 2.5 \times 0.45 = 1.125 \;mol$

M(Na₃PO₄) = 163.94 g/mol

n(Na₃PO₄) $= \frac{86.2}{163.94} = 0.525 \;mol$

According to the reaction:

n(AgNO₃) : n(Na₃PO₄) = 3 : 1

Real quantities:

n(AgNO₃) : n(Na₃PO₄) = 1.125 : 0.525 = 2.14 : 1

So, silver nitrate is a limiting reagent.

B) n_1(Na₃PO₄) $= \frac{1}{3}n(AgNO_3) = \frac{1}{3} \times 1.125 = 0.375 \;mol$ (used in reaction)

Δn(Na₃PO₄) = 0.525 – 0.375 = 0.15 mol (the excess will remain)

C) n(Ag₃PO₄) $= \frac{1}{3}n(AgNO3) = \frac{1}{3} \times 1.125 = 0.375 \;mol$

M(Ag₃PO₄ ) = 418.58 g/mol

m(Ag₃PO₄) $= 0.375 \times 418.58 = 156.96 \;g$