85.65 grams of barium hydroxide were dissolved in 2000ml of water. 88 grams of carbon dioxide were passed through the solution producing a white precipitate. what was this white precipitate and how many grams of it were produced.
Ba(OH)2(aq) + CO2(g) ==> BaCO3(s) + H2O(l)
Determine the limiting reagent (it appears to be Ba(OH)2 but you should confirm that) and convert mols Ba(OH)2 to moles BaCO3 then to grams BaCO3.
Can you explain it a little more cuz i got so confused!!!
i had a question just like this so if you could explain it a little more it migth be helful for me as well.
Thank you!
To determine the identity of the white precipitate and calculate its mass, we need to understand the reaction that occurred between barium hydroxide (Ba(OH)2) and carbon dioxide (CO2).
The reaction between barium hydroxide and carbon dioxide leads to the formation of barium carbonate (BaCO3) and water (H2O). The balanced chemical equation for this reaction is:
Ba(OH)2 + CO2 -> BaCO3 + H2O
From the equation, we can see that one molecule of barium hydroxide reacts with one molecule of carbon dioxide to produce one molecule of barium carbonate and one molecule of water.
First, let's calculate the number of moles of barium hydroxide (Ba(OH)2) using its given mass of 85.65 grams. We can use the formula:
Number of moles = Mass / Molar mass
The molar mass of barium hydroxide (Ba(OH)2) can be calculated as follows:
Ba: 1 atom x atomic mass of Ba
O: 2 atoms x atomic mass of O
H: 2 atoms x atomic mass of H
Molar mass of Ba(OH)2 = (1 x atomic mass of Ba) + (2 x atomic mass of O) + (2 x atomic mass of H)
Next, let's calculate the number of moles of carbon dioxide (CO2) using its given mass of 88 grams. We can use a similar formula:
Number of moles = Mass / Molar mass
The molar mass of carbon dioxide (CO2) can be calculated as follows:
C: 1 atom x atomic mass of C
O: 2 atoms x atomic mass of O
Molar mass of CO2 = (1 x atomic mass of C) + (2 x atomic mass of O)
Now that we have calculated the number of moles of barium hydroxide and carbon dioxide, we can determine the limiting reactant. The substance that is entirely consumed in the reaction is called the limiting reactant, and the product formed from it determines the mass of the precipitate.
To calculate the limiting reactant, we compare the moles of each reactant. Whichever reactant has the smaller number of moles will be the limiting reactant.
To determine the mass of the precipitate (barium carbonate), we need to use the stoichiometry relationship between the limiting reactant (carbon dioxide) and the product (barium carbonate). From the balanced chemical equation, the stoichiometric ratio between carbon dioxide and barium carbonate is 1:1.
Therefore, the mass of the precipitate (barium carbonate) can be calculated using the formula:
Mass = Moles of limiting reactant (carbon dioxide) x Molar mass of barium carbonate
The molar mass of barium carbonate (BaCO3) can be calculated as follows:
Ba: 1 atom x atomic mass of Ba
C: 1 atom x atomic mass of C
O: 3 atoms x atomic mass of O
Molar mass of BaCO3 = (1 x atomic mass of Ba) + (1 x atomic mass of C) + (3 x atomic mass of O)
By performing each of the above calculations, we can determine the identity of the white precipitate and the mass of it that was produced.