A 10.0 g sample of a mixture of calcium chloride and sodium chloride is treated with Na2CO3 to precipitate the calcium as calcium carbonate. This CaCO3 is heated to convert all the calcium to CaO and the final mass of CaO is 1.62 gms. The % by mass of CaCl2 in the original mixture is:
Correct answer: B. 32.1%
- A. 15.2%
- B. 32.1%
- C. 21.8%
- D. 11.07%
Explanation
The problem involves determining the percentage by mass of CaCl2 in a mixture. First, calculate the moles of CaO formed by dividing the mass of CaO (1.62 g) by its molar mass (56.08 g/mol), resulting in 0.0289 moles. Since the conversion of CaCO3 to CaO is a 1:1 process, these moles represent the moles of CaCl2 in the original mixture. Multiply these moles by the molar mass of CaCl2 (110.98 g/mol) to find the mass of CaCl2, which is 3.21 g. The percentage by mass is then calculated by dividing the mass of CaCl2 by the total mass of the mixture (10.0 g) and multiplying by 100, yielding approximately 32.1%. Other options do not follow the correct stoichiometric calculations, leading to incorrect percentages.
Last updated
About Electrochemistry
Electrochemistry describes electron transfer through oxidation and reduction, identifies oxidizing and reducing agents, and applies oxidation numbers to balance redox equations. It also covers electrode potential and the standard hydrogen electrode, which provides the reference for comparing the reduction tendencies of other electrodes.
Practise Electrochemistry
696 free Electrochemistry MCQs from Chemistry, each with the correct answer and an explanation. Unlimited attempts, no account needed.
Exams that ask Chemistry questions like this
Chemistry is on 12 papers prepared for on TestUstad, and all of them draw the same bank, so this question is worth knowing for every one of them.
Related questions
10-2 mole of Fe3O4 is treated with excess KI solution in presence of dilute H2SO4, the products are Fe2+ and I2(g). What volume of 0.1 (M) Na2S2O3 will be needed to reduce the liberated I2(g)?
+3 and +5 oxidation states are often shown by:
3 faraday of electricity is passed through molten Al2O3 , aqueous solution of CuSO4 and molten NaCl taken in three different electrolytic cells. The amount of Al, Cu and Na deposited at the cathodes will be in the ratio of:
5.3 gm of M2CO3 is dissolved in 150 ml of 1N HCl. The unused acid required 100 ml of 0.5N NaOH. Hence equivalent weight of M is:
8H++ MnO4- → Mn2+ + 4H2O , which one is correct about given equation?