answersLogoWhite

0


Best Answer

Add a reflux agent (CaF2) and electrolyse in a cell.

User Avatar

Wiki User

12y ago
This answer is:
User Avatar
More answers
User Avatar

AnswerBot

6mo ago

You can convert CaCO3 (calcium carbonate) to Ca (calcium) by heating it at a high temperature in the presence of a reducing agent, such as carbon. This process, known as calcination, will decompose the calcium carbonate, releasing carbon dioxide and leaving behind calcium oxide. The calcium oxide can then be reacted with water to form calcium hydroxide, which can further be processed to obtain pure calcium.

This answer is:
User Avatar

User Avatar

Wiki User

12y ago

By heating it :)

This answer is:
User Avatar

User Avatar

Wiki User

15y ago

How can you convert CaCO3 to Ca?

This answer is:
User Avatar

Add your answer:

Earn +20 pts
Q: How can you convert CaCO3 to Ca?
Write your answer...
Submit
Still have questions?
magnify glass
imp
Related questions

How many Ca atoms are found in 0.50 moles of CaCO3?

0.50 moles CaCO3 (1 mole Ca/1 mole CaCO3)(6.022 X 1023/1 mole Ca)= 3.0 X 1023 atoms of calcium===================


What is the balanced equation for calcium hydroxide and carbon dioxide produce calcium carbonate and water?

If you are after the symbol equation it's: Ca(OH)2(aq) + CO2 --> CaCO3 The word equation is: Calcium Hydroxide + Carbon Dioxide --> Calcium Carbonate (Limewater) (Limestone)


What is the balanced equation for the reaction of calcium and carbonic acid?

The balanced equation for the reaction of calcium (Ca) with carbonic acid (H2CO3) is: Ca + H2CO3 -> CaCO3 + H2


What does Ca and HCO3 make?

CaCO3, which is calcium carbonate.


What is the mass percent of calcium in calcium suppleiment if 1.15g of CaCO3 is extracted from a 2.70g pill?

First, calculate the mass of calcium in CaCO3: m(Ca) = (molar mass of Ca / molar mass of CaCO3) x mass of CaCO3. Then, calculate the mass percent of calcium: (mass of Ca / total mass of supplement) x 100. Given that molar mass of Ca = 40.08 g/mol and molar mass of CaCO3 = 100.09 g/mol, the mass percent of calcium will be (40.08 / 100.09) x 1.15 / 2.70 x 100.


How many grams of calcium in 34.5 of Ca CO 3?

To calculate the amount of calcium in 34.5 g of CaCO3, you need to consider the molar mass of CaCO3 which is 100.09 g/mol. Calcium accounts for approximately 40.08 g in every 100.09 g of CaCO3, which means there are (40.08/100.09) * 34.5 g = 13.82 g of calcium in 34.5 g of CaCO3.


How do you convert alkalinity as HCO3 to CaCO3?

To convert alkalinity (HCO3) to CaCO3, you need to use the molar mass ratio. For every mole of bicarbonate (HCO3), you have one mole of carbonate (CO3) in CaCO3. So, to convert, you can multiply the HCO3 concentration by a factor of 50.04 (molar mass of CaCO3/molar mass of HCO3).


What is the chemical formula for ca and co3?

The chemical formula for calcium is Ca and the chemical formula for carbonate is CO3. When combined, they form calcium carbonate with the chemical formula CaCO3.


What is the complete equation for CaCO3 plus 2H2O?

The chemical reaction between CaCO3 and 2H2O results in the formation of Ca(OH)2 (calcium hydroxide) and CO2 (carbon dioxide). The balanced equation for this reaction is: CaCO3 + 2H2O -> Ca(OH)2 + CO2.


What is the equation for the reaction of carbonic acid and calcium hydroxide?

The reaction between carbonic acid (H2CO3) and calcium hydroxide (Ca(OH)2) forms calcium carbonate (CaCO3) and water (H2O). The balanced chemical equation for the reaction is: H2CO3 + Ca(OH)2 → CaCO3 + 2H2O.


What happens when CaCO3 is added to HNO3?

When CaCO3 is added to HNO3, a chemical reaction occurs where CaCO3 reacts with HNO3 to produce Ca(NO3)2, CO2, and H2O. This reaction is a double displacement reaction where the calcium ions in CaCO3 switch places with the nitrate ions in HNO3.


What is the net ionic equation of Na2CO3 plus CaCl2 equals CaCo3?

The net ionic equation for the reaction between Na2CO3 and CaCl2 to form CaCO3 is: 2Na+ + CO3^2- + Ca^2+ + 2Cl- → CaCO3(s) + 2Na+ + 2Cl-. This equation represents the ions that are involved in the reaction, excluding spectator ions.