16. (A) M and R
17. (A) 0.0067 moles
Explanation:
We have the following chemical reaction:
M₂R₃ (aq) + 3 TS₂ (aq) → 2 MS₃ (s) + 3 TR (s)
molar concentration = number of moles / volume
number of moles = molar concentration × volume
number of moles of M₂R₃ = 1 M × 20 mL = 20 mmoles
number of moles of TS₂ = 1 M × 50 mL = 50 mmoles
Taking the account the stoichiometry of the chemical reaction we devise the following reasoning:
if 1 mmole of M₂R₃ will react with 3 mmoles TS₂
then X mmoles of M₂R₃ will react with 50 mmoles TS₂
X = (1 × 50) / 3 = 16.67 mmoles of M₂R₃
We have 20 mmoles of M₂R₃ so in the final solution we will have 20-16.67 = 3.33 mmoles of unreacted M₂R₃.
16. After completion of the chemical reaction we have the M³⁺ and R²⁻ ions dissolved in solution.
17. After completion of the chemical reaction we have 3.33 mmoles of unreacted M₂R₃ in aqueous state.
if 1 mmole of M₂R₃ have 2 mmoles of M³⁺ ions and 3 mmoles of R²⁻ ions
then 3.33 mmoles of M₂R₃ have X mmoles of M³⁺ ions and Y mmoles of R²⁻ ions
X = (3.33 × 2) / 1 = 6.66 mmoles = 0.0067 moles of M³⁺ ions
Y = (3.33 × 3) / 1 = 9.99 mmoles = 0.0099 moles ≈ 0.01 moles of R²⁻ ions
The number of moles for the ion with the lowest concentration will be 0.0067 moles for M³⁺ ions.
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