What is the mass of calcium carbonate precipitated when calcium chloride solution is added to a solution containing 21.2g of sodium carbonate?
RMM of sodium carbonate = 106
RMM of calcium carbonate = 100
Asked by ben_kpg
| 20th Jun, 2019,
04:12: PM
Expert Answer:
NaCO3 + CaCl2 → NaCl + CaCO3
21.2g
The relative molar mass of NaCO3 is 106 g per mole.
This means mass of 1 mole of NaCO3 = 106 g
1 mole of NaCO3 liberates 1 mole of CaCO3
Therefore 0.2 mole of NaCO3 liberates 0.2 mole of CaCO3
= 100×0.2 .......................... (relative molar mass of CaCO3=100)
= 20g
NaCO3 + CaCl2 → NaCl + CaCO3
The relative molar mass of NaCO3 is 106 g per mole.
This means mass of 1 mole of NaCO3 = 106 g
Answered by Ramandeep
| 21st Jun, 2019,
02:07: PM
Concept Videos
- How to calculate Emphirical formula and Molecular formula? thanks.
- In Chemistry, in a given reaction , how to recognize which reaction it is i.e oxidation,reduction or redox reaction?
- b) one ..
- L
- Find the volume of oxygen at STP required for the complete combustion of 2 litres of carbon monoxide at S.T.P.
- why mole concept important
- A mixture of hydrogen and chlorine occupying 36 cm3 was exploded. On shaking it with water, 4cm3 of hydrogen was left behind. Find the composition of the mixture.
- What volume of oxygen at STP is required to affect the combustion of 11 litres of ethylene [C2H4] at 273o C and 380 mm of Hg pressure?
C2H4+3O2
2CO2 + 2H2O
- Calculate the number of molecules in 1 kg of Calcium Chloride.
- No. of moles in 32g of sulphur molecule.
Kindly Sign up for a personalised experience
- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions
Sign Up
Verify mobile number
Enter the OTP sent to your number
Change