URGENT:DUE TMRW(WILL GIVE BRAINLIEST TO FASTEST CORRECT ANSWER)
Use the following information to answer questions 16 and 17.
M2R3(aq) + 3TS2(aq) → 2MS3(s) + 3TR(S)
M2R3 and TS, are both ionic compounds. When 20.0 mL of a
1.0 M solution of M2R3 and 50.0 mL of a 1.0 M solution of TS2
are mixed, two precipitates are formed.
16. After the reaction has gone to completion, what ions remain in
solution?
(A) M and R
(B) T and S
(C) M and S
(D) T and R
17. After the reaction has gone to completion, what is the number
of moles of the ion that appears in the lowest concentration in
the solution?
(A) 0.0067 mole
(B) 0.010 mole
(C) 0.020 mole
(D) 0.060 mole​​

Respuesta :

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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