211k views
2 votes
What concentration of ClO3â results when 967 mL of 0.367 M AgClO3 is mixed with 667 mL of 0.643 M Mn(ClO3)2?

1 Answer

1 vote

Answer:

0.742M is the concentration of ClO₃⁻

Step-by-step explanation:

The AgClO₃ contains 1 mole of ClO₃⁻ per mole of compound, but Mn(ClO₃)₂ contains 2 moles of ClO₃⁻ per mole of compound.

The moles of AgClO₃ are:

0.967L * (0.367moles / L) = 0.355 moles AgClO₃ = 0.355 moles of ClO₃⁻

And of Mn(ClO₃)₂:

0.667L * (0.643moles / L) = 0.429 moles Mn(ClO₃) * 2 = 0.858 moles of ClO₃⁻

That means total moles of ClO₃⁻ are:

0.355 moles + 0.858 moles = 1.213 moles ClO₃⁻ that are in 0.967L + 0.667L = 1.634L.

The concentration is:

1.213 moles ClO₃⁻ / 1.634L =

0.742M is the concentration of ClO₃⁻

User Adam Thomason
by
7.2k points