Titration is a process where a known concentration solution is used for determining the concentration of unknown solution. Known concentration solution called as titrant while unknown concentrated solution called as analyte.
Mass percentage: mass percentage represents the purity of the component present in the mixture.
Molarity: molarity is the ratio of the number of moles to the volume of the solution.
Therefore, moles of is used to react with excess present in given reaction.
The balanced chemical equation to the given acid-base titration reaction
According to this balanced chemical equation, 1 mole of reacts with 1mole of
So
Number of moles of needed for moles of = moles.
Number of moles of excess =mol
Number of moles of used to react with
Therefore, moles of is used in the reaction with
Given balanced chemical equation
According to the given balanced chemical equation, 1mole of reacts with 2moles of
Ans:The mass percentage of the in the sample is 74.4%
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 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.
< Question 1 of 3 > 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) + 2 HCl(aq) → CaCl, (aq) + H2O(1) + CO2(g) The excess HCl(aq) is titrated by 8.45 mL of 0.125 M NaOH(aq). Calculate the mass percentage of CaCO,(s) in the sample. mass percentage:
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...
Calcium carbonate (CaCO3) reacts with stomach acid (HCl, hydrochloric acid) according to the following equation: CaCO3(s)+2HCl(aq)⟶CO2(g)+H2O(l)+CaCl2(aq) Tums, an antacid, contains CaCO3. If Tums is added to 10.0 mL of a solution that is 0.400 M in HCl, how many grams of CO2 gas are produced?
Answer the following for the reaction: CaCO3(s)+2HCl(aq)→H2O(l)+CO2(g)+CaCl2(aq) What is the molarity of a HCl solution if the reaction of 215. mL of the HCl solution with excess CaCO3 produces 10.7 LL of CO2 gas at 725 mmHg and 18∘C?
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) was back titrated with 24.70 mL of 0.0980 M NaOH(aq). How many moles of HCl react with the carbonate? moles of HCI = mol What is the identity of the cation, X? cation: 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...
How many mL of 0.572 M HCl are needed to dissolve 9.65 g of CaCO3?2HCl(aq) +CaCO3(s) CaCl2(aq) + H2O(l) + CO2(g)________mL
Calcium carbonate (CaCO3) reacts with stomach acid (HCl, hydrochloric acid) according to the following equation: CaCO3(s) + 2HCl(aq) + CO2(g) + H2O(l) + CaCl2(aq) A typical antacid contains CaCO3. If such an antacid is added to 25.0 mL of a solution that is 0.300 M in HCl, how many grams of CO2 gas are produced? Express the mass to three significant figures and include the appropriate units. TI MÃ + + + a ? Value Units MCO: = Submit Previous...
How many milliliters of 0.480 M HCl are needed to react with 54.8 g of CaCO3? 2HCl(aq) + CaCO3(s) → CaCl2(aq) + CO2(g) + H2O(l) mL?
Calcium carbonate, CaCO3, reacts with stomach acid, (HCI, hydrochloric acid) according to the following equation: CaCO3(s) + 2HCl(aq)-CO2(g) + H2O(1) +CaCl2(aq) Tums, an antacid, contains CaCO3. If Tums is added to 20.0 mL of a 0.400 M HCl solution, how many grams of CO2 gas are produced?