Calculate the mass of solid which will be precipitated when 35g of carbon(iv)oxide are passed through a solution of calcium hydroxide

Ca(OH)2 + CO2 ==> CaCO3 + H2O

mols CO2 = grams/molar mass = ?
ASSUMING you have enough Ca(OH)2 to use all of the CO2, then
mols Ca(OH)2 = mols CO2 (look and see in the balanced equation that 1 mole Ca(OH)2 requires 1 mol CO2 and ppts 1 mol CaCO3.
Then convert mols CaCO3 to grams.
grams = mols x molar mass = ?
Post your work if you get stuck.

kindly how the workings

Calculate the grammes of oxygen which would be evolve on heating 5g of potassium trioxochlorate (v)

I can't understand anything,kindly show the workings

To calculate the mass of the solid precipitated when carbon dioxide (CO2) is passed through a solution of calcium hydroxide (Ca(OH)2), we need to determine the balanced chemical equation of the reaction between carbon dioxide and calcium hydroxide.

The balanced chemical equation for the reaction is:

CO2 + Ca(OH)2 → CaCO3 + H2O

From the balanced equation, we can see that one mole of CO2 reacts with one mole of Ca(OH)2 to produce one mole of CaCO3 and one mole of H2O.

Next, we need to determine the molar mass of CO2 and CaCO3.

The molar mass of CO2 = 12.01 g/mol (carbon) + (2 * 16.00 g/mol) (oxygen) = 44.01 g/mol
The molar mass of CaCO3 = 40.08 g/mol (calcium) + 12.01 g/mol (carbon) + (3 * 16.00 g/mol) (oxygen) = 100.09 g/mol

Now, let's calculate the number of moles of CO2:

Number of moles of CO2 = mass of CO2 / molar mass of CO2
= 35 g / 44.01 g/mol ≈ 0.795 mol

Since the reaction is 1:1 between CO2 and CaCO3, the number of moles of CaCO3 formed will also be 0.795 mol.

Finally, let's calculate the mass of CaCO3:

Mass of CaCO3 = number of moles of CaCO3 * molar mass of CaCO3
= 0.795 mol * 100.09 g/mol ≈ 79.5 g

Therefore, approximately 79.5 grams of calcium carbonate (CaCO3) will be precipitated when 35 grams of carbon dioxide (CO2) are passed through a solution of calcium hydroxide (Ca(OH)2).

Why did the carbon dioxide go to therapy? Because it was feeling a bit precipitated!

Now, to calculate the mass of the solid precipitate formed when 35g of carbon dioxide (CO2) reacts with calcium hydroxide (Ca(OH)2), we need to know the balanced chemical equation for the reaction.

The balanced equation is:

CO2 + Ca(OH)2 -> CaCO3 + H2O

From the equation, we can see that 1 mole of CO2 reacts with 1 mole of Ca(OH)2 to produce 1 mole of CaCO3.

To find the molar mass of CaCO3, we add up the atomic masses of calcium (Ca), carbon (C), and three oxygen (O) atoms:

CaCO3 = (40.08 g/mol + 12.01 g/mol + 3 * 16.00 g/mol) = 100.09 g/mol

Now, let's calculate the number of moles of CO2 in 35g of CO2:

moles = mass / molar mass
moles = 35g / 44.01 g/mol (molar mass of CO2)
moles ≈ 0.795 mol

Since the balanced equation shows a 1:1 ratio between CO2 and CaCO3, we know that 0.795 moles of CO2 will react to form the same number of moles of CaCO3.

Now, let's calculate the mass of CaCO3 formed:

mass = moles * molar mass
mass = 0.795 mol * 100.09 g/mol (molar mass of CaCO3)
mass ≈ 79.59 g

Therefore, approximately 79.59 grams of solid calcium carbonate (CaCO3) will be precipitated when 35 grams of carbon dioxide are passed through a solution of calcium hydroxide. Enjoy the chemistry show!