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What happens when you mix sodium percarbonate, sodium hydroxide and water?
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Keith Roman
What happens when you mix sodium percarbonate, sodium hydroxide and water?
Not much. Sodium per carbonate is a complex formed from sodium carbonate (Na2CO3) and hydrogen peroxide (H2O2). The formula is (Na2CO3)2:(H2O2)3. When you dissolve this in water you get a solution of sodium carbonate and hydrogen peroxide.
(Na2CO3)2:(H2O2)3(s) + H2O → 2 Na2CO3 + 3 H2O2
OK, so add some sodium hydroxide to this and it will become more basic, but that is about it.
Not much. Sodium per carbonate is a complex formed from sodium carbonate (Na2CO3) and hydrogen peroxide (H2O2). The formula is (Na2CO3)2:(H2O2)3. When you dissolve this in water you get a solution of sodium carbonate and hydrogen peroxide.
(Na2CO3)2:(H2O2)3(s) + H2O → 2 Na2CO3 + 3 H2O2
OK, so add some sodium hydroxide to this and it will become more basic, but that is about it.
It does, but in TINY quantities. When NaCl dissolves in water it dissociates into Na+ ions, Cl- ions. The water is already partly dissociated into H2O, H+ and OH-. A few of the OH- ions will merge with the Na+ ions in the solution. Energetically they much prefer to be in their ionic states, but a scant few will exist as NaOH molecules. What a lot of people forget is that there are VERY FEW reactions that proceed in one direction and stay there. The vast majority exist in equilibrium between the reactants and the products. Such is the case here, but as I said, energetically they much prefer to just be ions.
It does, but in TINY quantities. When NaCl dissolves in water it dissociates into Na+ ions, Cl- ions. The water is already partly dissociated into H2O, H+ and OH-. A few of the OH- ions will merge with the Na+ ions in the solution. Energetically they much prefer to be in their ionic states, but a scant few will exist as NaOH molecules. What a lot of people forget is that there are VERY FEW reactions that proceed in one direction and stay there. The vast majority exist in equilibrium between the reactants and the products. Such is the case here, but as I said, energetically they much prefer to just be ions.
This gives info on a chemical reaction, then gives an amount of one substance and asks for an amount of a different substance. That's how you identify a stoichiometry question.
The steps in a stoichiometry question are:
Write a balanced chemical equation for the reaction.
Convert the given amount of substance to moles of that substance.
Multiply by a mole ratio of the form: (coefficient of wanted substance from balanced equation in moles of that substance/ coefficient of given substance from balanced equation in moles of that substance) to give moles of wanted substance.
Convert moles of wanted substance to desired units.
Reaction info given: sodium oxide reacts with water to make sodium hydroxide.
Using rules for writing formulas, sodium oxide Na+ and O2- so Na2O.
Water H2O
Sodium hydroxide: Na+ and OH- so NaOH.
Then unbalanced equation:
Na2O + H2O → NaOH
Fix Na first
Na2O + H2O → 2NaOH.
That also fixed both the hydrogen and oxygen. So it's balanced.
Step 2: convert to moles
140 g NaOH x (1 mole NaOH/ 40.00 g NaOH) = # moles NaOH
This gives info on a chemical reaction, then gives an amount of one substance and asks for an amount of a different substance. That's how you identify a stoichiometry question.
The steps in a stoichiometry question are:
Write a balanced chemical equation for the reaction.
Convert the given amount of substance to moles of that substance.
Multiply by a mole ratio of the form: (coefficient of wanted substance from balanced equation in moles of that substance/ coefficient of given substance from balanced equation in moles of that substance) to give moles of wanted substance.
Convert moles of wanted substance to desired units.
Reaction info given: sodium oxide reacts with water to make sodium hydroxide.
Using rules for writing formulas, sodium oxide Na+ and O2- so Na2O.
Water H2O
Sodium hydroxide: Na+ and OH- so NaOH.
Then unbalanced equation:
Na2O + H2O → NaOH
Fix Na first
Na2O + H2O → 2NaOH.
That also fixed both the hydrogen and oxygen. So it's balanced.
Step 2: convert to moles
140 g NaOH x (1 mole NaOH/ 40.00 g NaOH) = # moles NaOH
Not much. Sodium per carbonate is a complex formed from sodium carbonate (Na2CO3) and hydrogen peroxide (H2O2). The formula is (Na2CO3)2:(H2O2)3. When you dissolve this in water you get a solution of sodium carbonate and hydrogen peroxide.
(Na2CO3)2:(H2O2)3(s) + H2O → 2 Na2CO3 + 3 H2O2
OK, so add some sodium hydroxide to this and it will become more basic, but that is about it.
Not much. Sodium per carbonate is a complex formed from sodium carbonate (Na2CO3) and hydrogen peroxide (H2O2). The formula is (Na2CO3)2:(H2O2)3. When you dissolve this in water you get a solution of sodium carbonate and hydrogen peroxide.
(Na2CO3)2:(H2O2)3(s) + H2O → 2 Na2CO3 + 3 H2O2
OK, so add some sodium hydroxide to this and it will become more basic, but that is about it.
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It does, but in TINY quantities. When NaCl dissolves in water it dissociates into Na+ ions, Cl- ions. The water is already partly dissociated into H2O, H+ and OH-. A few of the OH- ions will merge with the Na+ ions in the solution. Energetically they much prefer to be in their ionic states, but a scant few will exist as NaOH molecules. What a lot of people forget is that there are VERY FEW reactions that proceed in one direction and stay there. The vast majority exist in equilibrium between the reactants and the products. Such is the case here, but as I said, energetically they much prefer to just be ions.
It does, but in TINY quantities. When NaCl dissolves in water it dissociates into Na+ ions, Cl- ions. The water is already partly dissociated into H2O, H+ and OH-. A few of the OH- ions will merge with the Na+ ions in the solution. Energetically they much prefer to be in their ionic states, but a scant few will exist as NaOH molecules. What a lot of people forget is that there are VERY FEW reactions that proceed in one direction and stay there. The vast majority exist in equilibrium between the reactants and the products. Such is the case here, but as I said, energetically they much prefer to just be ions.
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This gives info on a chemical reaction, then gives an amount of one substance and asks for an amount of a different substance. That's how you identify a stoichiometry question.
The steps in a stoichiometry question are:
Reaction info given: sodium oxide reacts with water to make sodium hydroxide.
Using rules for writing formulas, sodium oxide Na+ and O2- so Na2O.
Water H2O
Sodium hydroxide: Na+ and OH- so NaOH.
Then unbalanced equation:
Na2O + H2O → NaOH
Fix Na first
Na2O + H2O → 2NaOH.
That also fixed both the hydrogen and oxygen. So it's balanced.
Step 2: convert to moles
140 g NaOH x (1 mole NaOH/ 40.00 g NaOH) = # moles NaOH
You fill in the number, #
Step 3: moles ratio
# moles NaOH x (1 mole Na2O/ 2 moles NaOH) = ## moles Na2O
You fill in number if moles, ##
Step 4: convert to desired units
## moles Na2O x (61.98 g Na2O/ 1 moles Na2O) = ### g Na2O, your answer.
This gives info on a chemical reaction, then gives an amount of one substance and asks for an amount of a different substance. That's how you identify a stoichiometry question.
The steps in a stoichiometry question are:
Reaction info given: sodium oxide reacts with water to make sodium hydroxide.
Using rules for writing formulas, sodium oxide Na+ and O2- so Na2O.
Water H2O
Sodium hydroxide: Na+ and OH- so NaOH.
Then unbalanced equation:
Na2O + H2O → NaOH
Fix Na first
Na2O + H2O → 2NaOH.
That also fixed both the hydrogen and oxygen. So it's balanced.
Step 2: convert to moles
140 g NaOH x (1 mole NaOH/ 40.00 g NaOH) = # moles NaOH
You fill in the number, #
Step 3: moles ratio
# moles NaOH x (1 mole Na2O/ 2 moles NaOH) = ## moles Na2O
You fill in number if moles, ##
Step 4: convert to desired units
## moles Na2O x (61.98 g Na2O/ 1 moles Na2O) = ### g Na2O, your answer.
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