A 0.6739 g sample of a pure carbonate, X,CO,(s), was dissolved in 50.0 mL of 0.1850 M HCl(aq). The excess HCl(aq)...

Question

Question

Answer 1

Answer #1

Question 1.

(i) Moles of HCl react with the carbonate = {(50/1000) L * 0.185 mol/L} - {(24.7/1000) L * 0.098 mol/L} = 0.0068294 mol

(ii) The identity of the cation (X): Molar mass of X_nCO₃ = 0.6739 g/(0.0068294 mol) = 98.676 g/mol; Cation (X) = Ca

Note: The molar mass of CaCO₃ = 100 g/mol

Add a comment

Answer 2

Similar Homework Help Questions

A standardized solution that is 0.0100 M in Na+ is necessary for a Hame photometric determination...

A standardized solution that is 0.0100 M in Na+ is necessary for a Hame photometric determination of the element. How many grams of primary-standard-grade sodium carbonate are necessary to prepare 700.0 mL of this solution? Calculate the number of moles of solute in 79.21 mL of 0.1355 M K, Cr, o, (aq). moles of solute: mol
please help! A 0. 563 g sample of a pure carbonate, X_nCO_3(s) was dissolved in 50....

please help! A 0. 563 g sample of a pure carbonate, X_nCO_3(s) was dissolved in 50. 0 mL of 0. 2000 M HCl(aq). The excess HCl(aq) was back titrated with 24. 20 mL of 0. 0980 M NaOH(aq). How many moles of HCl react with the carbonate? What is the identity of the cation, X?
A 0.450 g sample of impure CaCO3(s) is dissolved in 50.0 mL of 0.150 M HCl(aq)...

A 0.450 g sample of impure CaCO3(s) is dissolved in 50.0 mL of 0.150 M HCl(aq) . The equation for the reaction is CaCO3(s)+2HCl(aq)⟶CaCl2(aq)+H2O(l)+CO2(g) The excess HCl(aq) is titrated by 9.05 mL of 0.125 M NaOH(aq) . Calculate the mass percentage of CaCO3(s) in the sample.

A 0.450 g sample of impure CaCO, (s) is dissolved in 50.0 mL of 0.150 M...

A 0.450 g sample of impure CaCO, (s) is dissolved in 50.0 mL of 0.150 M HCl(aq). The equation for the reaction is CaCO,(s) + 2 HCl(aq) — CaCl, (aq) + H, 0(1) + CO2(g) The excess HCl(aq) is titrated by 4.85 mL of 0.125 M NaOH(aq) Calculate the mass percentage of Caco, (s) in the sample. mass percentage: The zinc content of a 1.03 g ore sample was determined by dissolving the ore in HCl, which reacts with the...
A 0.450 gram sample of impure CaCO3(s) is dissolved in 50.0 mL of 0.150 M HCl(aq). The equation for the reaction is Ca...

A 0.450 gram sample of impure CaCO3(s) is dissolved in 50.0 mL of 0.150 M HCl(aq). The equation for the reaction is CaCO3(s) + 2HCl(aq) arrow CaCl2(aq) +H2O(l) +CO2(g) The excess HCL (aq) is titrated by 6.45 mL of 0.125 M NaOH(aq). Calculate the mass percentaageof CaCO3(s) in the sample. Please show steps.
A 50.0 mL sample of 0.0639 M AgNO3(aq) is added to 50.0 mL of 0.100 M...

A 50.0 mL sample of 0.0639 M AgNO3(aq) is added to 50.0 mL of 0.100 M Naloz(aq). Calculate the (Ag") at equilibrium in the resulting solution. The Kp value for AglO3(s) is 3.17 x 10-8 (Ag"]= [ mol/l

If 50.0 g of CH3OH (MM = 32.04 g/mol) are dissolved in 500.0 mL of solvent

If 50.0 g of CH3OH (MM = 32.04 g/mol) are dissolved in 500.0 mL of solvent, what is the concentration of CH3OH in the resulting solution? How many moles of CH3OH are there in 50.0 mL of 0.400 M CH3OH? How many mL of 0.300 M NaCI solution are required to produce 0.150 moles of NaCI?
A 50.0-mL sample of 0.15 M butanoic acid, CH3CH2CH2COOH, is titrated with 0.30 M NaOH(aq). Ka...

A 50.0-mL sample of 0.15 M butanoic acid, CH3CH2CH2COOH, is titrated with 0.30 M NaOH(aq). Ka for butanoic acid is 1.52 x 10-5 a) How many mL of NaOH(aq) are required to reach the equivalence point? b)What is the pH of the solution after 27.0 mL of NaOH(aq) have been added?
A 0.1983 g sample of Na2CO3 was dissolved in 100.00 ml H20. That solution was titrated with 0.1531 M HCL as titrant. The...

A 0.1983 g sample of Na2CO3 was dissolved in 100.00 ml H20. That solution was titrated with 0.1531 M HCL as titrant. The PKa1 and PKa2 of carbonic acid (the conjugate acid of the carbonate ion) are 6.351 and 10.329, respectively. 1.) What is the initial pH of the sodium carbonate solution. 2.) What is the first equivalence point (in ml HCL)? What is the pH at 1st eq.? 3.) What is the second equivalence point (in ml HCL)? What...

Data and Results: Part A M (HCI): Volume HCl(aq): Reaction 2: Heat of Neutralization M(NaOH) 50.0...

Data and Results: Part A M (HCI): Volume HCl(aq): Reaction 2: Heat of Neutralization M(NaOH) 50.0 mL (+0.5 mL) Volume NaOH(aq): 20, 9 °C Final Temperature: 6.7 (not from the data above) 1.0 50-0 ML (+0.5 mL) 27.6 °C Initial Temperature: AT (from plot) Mass of Solution: (show all calculations) Moles H*:(show all calculations) Moles OH :(show all calculations) 0.05 moles H 0.05 moles OH AH experimental: (show all calculations) kJ/mol AH (theoretical): (see page 8) (show all calculations) kJ/mol...