95.65 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.
explain please!!!! i got so confused with this question!!!!
i really need help with this right now!!!!
The ppt is BaCO3
This is a limiting reagent problem. I know that because amounts are given for BOTH reactants.
Ba(OH)2 + CO2 ==> BaCO3 + H2O
1. Convert 95.65g Ba(OH)2 to mols. mols = grams/molar mass
2. Do the same for 88 g CO2.
3a. Using the coefficients in the balanced equation, convert mols Ba(OH)2 to mols BaCO3.
3b. Do the same for mols CO2.
3c. You will have two values for mols BaCO3 produced and only one can be correct. The correct value in limiting reagent problems is ALWAYS the smaller one and the reagent producing that number is the limiting reagent.
4. Using the smaller number, convert mols to g. g BaCO3 = mols x molar mass BaCO3.
To determine the white precipitate formed and the amount of it produced, we need to understand the chemical reaction that occurred when carbon dioxide was passed through the barium hydroxide solution.
The balanced chemical equation for this reaction can be written as follows:
Ba(OH)2 + CO2 -> BaCO3 + H2O
From the equation, we can see that when barium hydroxide (Ba(OH)2) reacts with carbon dioxide (CO2), it forms barium carbonate (BaCO3) and water (H2O).
Now, let's calculate the molar mass of barium hydroxide (Ba(OH)2) and carbon dioxide (CO2).
The molar mass of barium (Ba) = 137.33 g/mol
The molar mass of oxygen (O) = 16.00 g/mol
The molar mass of hydrogen (H) = 1.01 g/mol
Molar mass of barium hydroxide (Ba(OH)2) = (1 x 137.33) + (2 x 16.00) + (2 x 1.01) = 171.34 g/mol
Molar mass of carbon dioxide (CO2) = (1 x 12.01) + (2 x 16.00) = 44.01 g/mol
We know that 95.65 grams of barium hydroxide (Ba(OH)2) were dissolved in 2000 ml of water. To find the number of moles of Ba(OH)2 dissolved, we'll use the formula:
Number of moles = mass / molar mass
Number of moles of Ba(OH)2 = 95.65 g / 171.34 g/mol
Next, we can use the balanced chemical equation to calculate the stoichiometry between Ba(OH)2 and BaCO3. From the equation, we can see that the ratio between Ba(OH)2 and BaCO3 is 1:1. This means that every 1 mole of Ba(OH)2 reacts to produce 1 mole of BaCO3.
So, the number of moles of BaCO3 produced will be the same as the number of moles of Ba(OH)2 dissolved.
Now, let's calculate the mass of BaCO3 produced using the formula:
Mass = number of moles x molar mass
Mass of BaCO3 = Number of moles of BaCO3 x molar mass of BaCO3
Finally, we can now determine the white precipitate formed (BaCO3) and the amount of it produced in grams.
I hope this explanation guides you in solving the problem.