Dumbledore'sArmy
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#1
Report Thread starter 6 years ago
#1
Please help ke with this...its really hard

I did this experiment

Stage 1: Accurately weigh 4.80 of copper(II) nitrate trihydrate, Cu(NO3)2.3H2O into a beaker and dissolve in 20cm3 deionised water. Slowly add 30cm3 of 2 mol dm-3sodium hydroxide solution (NaOH), swirling your beaker continuously.

What is the balanced chemical equation?

Stage 2: Add a further 100-3 deionised water to the percipitate of copper(II) hydroxide in your beaker.

What was the purpose of adding 100-3 deionised water?

And finally, what could i do, if I was doing this reaction, to maximise the cost effectiveness of the reaction, since i found that this reaction is 63.35% in atom economy.

Please help, I attempted this, and its all chaos at the moment
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Dumbledore'sArmy
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#2
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Stage 1: Accurately weigh 4.80 of copper(II) nitrate trihydrate, Cu(NO3)2.3H2O into a beaker and dissolve in 20cm3 deionised water. Slowly add 30cm3 of 2 mol dm-3sodium hydroxide solution (NaOH), swirling your beaker continuously.

What is the balanced chemical equation?

Stage 2: Add a further 100-3 deionised water to the percipitate of copper(II) hydroxide in your beaker.

What was the purpose of adding 100-3 deionised water?

And finally, what could i do, if I was doing this reaction, to maximise the cost effectiveness of the reaction, since i found that this reaction is 63.35% in atom economy.

I dont really understand the balancing equation part (even though i know how to balance) and the reason on adding water again.
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charco
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(Original post by Dumbledore'sArmy)
Stage 1: Accurately weigh 4.80 of copper(II) nitrate trihydrate, Cu(NO3)2.3H2O into a beaker and dissolve in 20cm3 deionised water. Slowly add 30cm3 of 2 mol dm-3sodium hydroxide solution (NaOH), swirling your beaker continuously.

What is the balanced chemical equation?

Stage 2: Add a further 100-3 deionised water to the percipitate of copper(II) hydroxide in your beaker.

What was the purpose of adding 100-3 deionised water?

And finally, what could i do, if I was doing this reaction, to maximise the cost effectiveness of the reaction, since i found that this reaction is 63.35% in atom economy.

I dont really understand the balancing equation part (even though i know how to balance) and the reason on adding water again.
You add 0.02 mol of copper(II) nitrate to 0.06 mol of sodium hydroxide so you are attempting to precipitate out all of the copper(II) hydroxide. This leaves sodium nitrate in solution.

To get to the balanced equation you must write out the formulae of the compounds reacting on the left and the formulae of the compounds produced on the right.

Then you balance by adding coefficients in front of each formula.

The extra water seems to be just to dilute the excess sodium hydroxide so as to make the copper(II) hydroxide more dispersed and effectively 'washed' of strong base before collection by filtration.

.... and please don't double post!
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Dumbledore'sArmy
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(Original post by charco)
You add 0.02 mol of copper(II) nitrate to 0.06 mol of sodium hydroxide so you are attempting to precipitate out all of the copper(II) hydroxide. This leaves sodium nitrate in solution.

To get to the balanced equation you must write out the formulae of the compounds reacting on the left and the formulae of the compounds produced on the right.

Then you balance by adding coefficients in front of each formula.

The extra water seems to be just to dilute the excess sodium hydroxide so as to make the copper(II) hydroxide more dispersed and effectively 'washed' of strong base before collection by filtration.

.... and please don't double post!
Sorry for double posting. I was becoming desperate.

Is this the equation:

Cu(NO3) + NaOH ----> Cu(OH)2 .......OR Cu(NO3). 3H2O + NaOH ---> Cu(OH)2 + NaNO3 + H2O ? I'll balance these myself.

EDIT: Im trying to make copper(II) hydroxide, Cu(OH)2...also what could I do to maximise the cost effectiveness of the reaction?
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charco
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(Original post by Dumbledore'sArmy)
Sorry for double posting. I was becoming desperate.

Is this the equation:

Cu(NO3) + NaOH ----> Cu(OH)2 .......
You have the formula for copper(II) nitrate incorrect and you are missing the sodium nitrate.

You NEVER include water of crystallisation in aqueous reaction formulae.
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Dumbledore'sArmy
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#6
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(Original post by charco)
You have the formula for copper(II) nitrate incorrect and you are missing the sodium nitrate.

You NEVER include water of crystallisation in aqueous reaction formulae.
therefore its this one: Cu(NO3). 3H2O + NaOH ---> Cu(OH)2 + NaNO3 + H2O ....without the Water (3H20)
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charco
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(Original post by Dumbledore'sArmy)
therefore its this one: Cu(NO3). 3H2O + NaOH ---> Cu(OH)2 + NaNO3 + H2O ....without the Water (3H20)
I just said you NEVER include the water of crystallisation ... :confused:
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Dumbledore'sArmy
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#8
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(Original post by charco)
I just said you NEVER include the water of crystallisation ... :confused:
I was making sure, sorry!

One last tiny question: If i was conducting this reaction, what might I do to maximise the cost effectiveness of the reaction, because my atom economy was 63.35?

Thanks for the help on that equation, youve been ever so helpful!
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