calculate the mass of solid which would be precipitated when 35g of carbon(IV)oxide are passed through a solution of calcium hydroxide
-monoxide or -dioxide ?
I think carbon(IV) oxide must be carbon dioxide.
CO2 + Ca(OH)2 ==> CaCO3(s) + H2O
mols CO2 = grams/molar mass = ?
1 mol CO2 = 1 mol CaCO3
grams CaCO3 = mols CaCO3 x molar mass CaCO3 = ?
Post your work if you get stuck.
To calculate the mass of solid precipitated when 35g of carbon dioxide (CO2) is passed through a solution of calcium hydroxide (Ca(OH)2), we need to use the balanced chemical equation for the reaction between CO2 and Ca(OH)2.
The balanced chemical equation is:
CO2 + Ca(OH)2 → CaCO3 + H2O
From the equation, we can see that one mole of CO2 reacts with one mole of Ca(OH)2 to form one mole of CaCO3.
Step 1: Find the molar mass of CO2.
The molar mass of CO2 is:
12.01 g/mol (C) + 16.00 g/mol (O) + 16.00 g/mol (O) = 44.01 g/mol
Step 2: Convert the mass of CO2 to moles.
Using the molar mass of CO2, we can convert the given mass to moles:
35 g CO2 * (1 mol CO2 / 44.01 g CO2) = 0.795 mol CO2
Step 3: Determine the molar ratio between CO2 and CaCO3.
From the balanced equation, the molar ratio between CO2 and CaCO3 is 1:1.
Step 4: Convert moles of CO2 to moles of CaCO3.
Since the molar ratio is 1:1, the moles of CaCO3 formed are also 0.795 mol.
Step 5: Find the molar mass of CaCO3.
The molar mass of CaCO3 is:
40.08 g/mol (Ca) + 12.01 g/mol (C) + 16.00 g/mol (O) + 16.00 g/mol (O) + 16.00 g/mol (O) = 100.09 g/mol
Step 6: Convert moles of CaCO3 to mass.
Using the molar mass of CaCO3, we can convert the moles to mass:
0.795 mol CaCO3 * (100.09 g CaCO3 / 1 mol CaCO3) = 79.57 g CaCO3
Therefore, the mass of solid precipitated when 35g of CO2 is passed through a solution of Ca(OH)2 is approximately 79.57 grams.
To calculate the mass of solid precipitate formed, we need to follow these steps:
Step 1: Write the balanced chemical equation for the reaction between carbon dioxide (CO2) and calcium hydroxide (Ca(OH)2).
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.
Step 2: Calculate the number of moles of CO2.
The molar mass of CO2 is 44 g/mol, so the number of moles of CO2 can be calculated using the equation:
moles = mass / molar mass
moles = 35 g / 44 g/mol
moles ≈ 0.795 moles
Step 3: Determine the stoichiometry of the reaction.
From the balanced equation, it is clear that 1 mole of CO2 reacts with 1 mole of Ca(OH)2 to produce 1 mole of CaCO3. Therefore, the number of moles of CaCO3 formed will be the same as the number of moles of CO2.
moles of CaCO3 formed = 0.795 moles
Step 4: Calculate the mass of CaCO3 using its molar mass.
The molar mass of CaCO3 is approximately 100 g/mol. The mass of the precipitated CaCO3 can be calculated using the equation:
mass = moles × molar mass
mass = 0.795 moles × 100 g/mol
mass ≈ 79.5 g
Therefore, approximately 79.5 grams of solid CaCO3 will be precipitated when 35 grams of CO2 are passed through a solution of calcium hydroxide.