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Dafna1 [17]
3 years ago
14

A 0.533 g sample of a carbonate salt is added to a flask. The only thing you know is that per formula unit, there is only one ca

rbonate anion, but the identity of the cation (or cations) is unknown. To determine the identity of the cations, 15.0 mL of 1.00 M sulfuric acid is added to the flask. After the resulting reaction completes, excess acid remains. The remaining solution is titrated with 0.5905 M NaOH(aq) and reaches the equivalence point when 29.393 mL of base is added. What is the identity of the compound?
Chemistry
1 answer:
dimulka [17.4K]3 years ago
6 0

Answer:

\mathbf{CrCO_3}

Explanation:

The total mole of H_2SO_4 added = 15 × 1 × 10⁻³

= 15 × 10⁻³ mole

= 15 mmoles

the number of moles of NaOH added in order to neutralize the excess acid = 0.5905 ×  29.393 = 17.36 mmoles

the equation for the reaction is:

2NaOH + H_2SO_4 --------> Na_2SO_4 +2H_2O

i.e

2 moles of NaOH react with H_2SO_4

1 moles of  NaOH will react with 1/2  H_2SO_4

17.36 mmoles of NaOH = 1/2 × 17.36 mmoles of H_2SO_4

                                       = 8.68 mmoles

Number of moles of H_2SO_4 that react with MCO₃ = Total moles of H_2SO_4 added - moles of H_2SO_4 reacted with NaOH

= (15 - 8.68) mmoles

= 6.32 mmoles

H_2SO_4 + MCO_3  ------>   M_2SO_4   +    H_2O   +   CO_2

1 mole of H_2SO_4 react with 1 mole of MCO_3

6.32 mmoles of H_2SO_4 = 6.32 mmoles of MCO_3

number of moles of MCO_3 = 6.32 10⁻³ moles

mass of  MCO_3 (carbonate salt) = 0.533 g

molar mass of MCO_3  = (M+60)g/mol

We all know that ;

number of moles = mass/molar mass

Then:

6.32 10⁻³ = 0.533 / (M+60)

(M+60) = 0.533/ 6.32 10⁻³

M + 60 = 84.34

M = 84.34 - 60

M = 24.34

Thus the element with the atomic mass of 24.34 is Chromium

The chemical formula for the compound is :  \mathbf{CrCO_3}

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1. 44.11 g

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