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!!!!
2 answers
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.
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.
3 answers