During a reaction of ca(oh)2 with carbon dioxide gas, 20g of solid caco3 is formed. calculate the mass of the nahco3 needed to produce the amount of carbon dioxide gas responsible for forming the caco3 precipitation.
Ca(OH)2 + CO2 ==> CaCO3 + H2O
mols CaCO3 formed = grams/molar mass = estimated 0.2 mol.
NaHCO3 + HCl ==> NaCl + H2O + CO2
Looking at the equations, you had 0.2 mol CaCO3 formed, and that was produced by 0.2 mol CO2. From the lower equation, for form 0.2 mol CO2 you need 0.2 mol NaHCO3.
Convert mols NaHCO3 to grams NaHCO3.
That's g = mols x molar mass = ?
To find the mass of NaHCO3 needed, we need to first determine the molar quantity of CO2 responsible for forming the precipitated CaCO3.
The balanced chemical equation for the reaction is:
2NaHCO3 + Ca(OH)2 → CaCO3 + 2H2O + 2CO2
From the equation, we can see that 2 moles of NaHCO3 produce 2 moles of CO2.
To calculate the molar quantity of CO2, we'll use the molar mass of CaCO3.
1 mole of CaCO3 = 40.08g + 12.01g + (3 * 16.00g) = 100.09g
Given that 20g of CaCO3 is formed, we can calculate the molar quantity:
Moles of CaCO3 = Mass of CaCO3 / Molar mass of CaCO3
= 20g / 100.09g/mol
= 0.1999 mol ≈ 0.2 mol
Since 2 moles of NaHCO3 are required to produce 2 moles of CO2, we can equate the moles of CO2 produced to the moles of NaHCO3 required:
0.2 mol CO2 = 2 mol NaHCO3
To find the mass of NaHCO3, we can use the molar mass of NaHCO3:
1 mole of NaHCO3 = 22.99g + 1.01g + 12.01g + (3 * 16.00g) = 84.01g
Mass of NaHCO3 = Moles of NaHCO3 × Molar mass of NaHCO3
= 0.2 mol × 84.01g/mol
= 16.802g
Therefore, approximately 16.802 grams of NaHCO3 are needed to produce the quantity of carbon dioxide gas responsible for forming 20 grams of CaCO3.