Mass of calcium trioxocarbonate (4) used in caco3+2hcl=cacl2+co2+h2o when 0.1 mole of hydrochloric acid react with calcium trioxocarbonate (4)
I need a solution
There is no such name as calcium trioxocarbonate (4). I know you mean CaCO3. An accepatble IUPAC is calcium carbonate. Where did you get that name? From a teacher? from book? from an article? Please tell me where you came up with that name and I'll respond with how to work the problem.
I made a mistake
To determine the mass of calcium trioxocarbonate (4) used in the given reaction, you need to consider the balanced chemical equation and the stoichiometry of the reaction.
First, let's write the balanced chemical equation for the reaction:
CaCO3 + 2HCl → CaCl2 + CO2 + H2O
From the balanced equation, we can see that 1 mole of calcium trioxocarbonate (CaCO3) reacts with 2 moles of hydrochloric acid (2HCl) to produce 1 mole of calcium chloride (CaCl2), 1 mole of carbon dioxide (CO2), and 1 mole of water (H2O).
Given that 0.1 mole of hydrochloric acid (HCl) reacts with calcium trioxocarbonate (CaCO3), we can determine the mole ratio between hydrochloric acid and calcium trioxocarbonate as 2:1.
Therefore, 0.1 mole of HCl corresponds to 0.1/2 = 0.05 mole of CaCO3.
Now, to calculate the mass of CaCO3 used, we need to use its molar mass. The molar mass of CaCO3 can be calculated as follows:
Molar mass of CaCO3 = (1 x atomic mass of Ca) + (1 x atomic mass of C) + (3 x atomic mass of O)
Looking up the atomic masses:
Atomic mass of Ca = 40.08 g/mol
Atomic mass of C = 12.01 g/mol
Atomic mass of O = 16.00 g/mol
Plugging in the values:
Molar mass of CaCO3 = (1 x 40.08) + (1 x 12.01) + (3 x 16.00) = 100.09 g/mol
Finally, to determine the mass of CaCO3 used, we can multiply the molar mass by the number of moles:
Mass of CaCO3 = Molar mass of CaCO3 x Number of moles
= 100.09 g/mol x 0.05 mol
= 5.0045 g (rounded to four decimal places)
Therefore, the mass of calcium trioxocarbonate (4) used in the reaction is approximately 5.0045 grams.