[NeonNe]

Hi, new undergrad here. I was hoping someone could help me get the ball rolling with theoretical yield calculations I need to work out. Need to figure out the Lead compounds in the products and finding it a little confusing.

2 reactions - Start using 3g Pb₃O₄ (s) with 100cm³ HNO₃ (aq)

(1) Pb₃O₄ (s) + 4HNO₃ (aq) ----> PbO₂ (s) + 2Pb(NO₃)₂ (aq) + 2H₂O

The 2Pb(NO₃)₂ (aq) was then split into two 50cm³ samples. One 50cm³ portion was boiled off to yield lead nitrate crystals and the other was taken for the next reaction:

(2) 3Pb(NO₃)₂ (aq) + 2NaCO₃ (aq) + 2NaOH (aq) ---> Pb(OH)₂.2PbCO₃ (s) + 6NaNO₃ (aq)

Any hints or tips would be very much appreciated.

[Borek]

This is simple stoichimetry, hard to tell where your problem lies.

http://www.chembuddy.com/?left=balancing-stoichiometry&right=stoichiometric-calculations

[NeonNe]

Hi Borek, thanks for the reply.

My knowledge of stoichimetry is admittedly quite basic thus far. The double product in the first reaction and the fact one of them is split in half to provide for the next reaction throws me off in my calculations.

Basically im using 0.004 mol Pb₃O₄ which is the limiting reagent. Does this mean I get 0.004 mol PbO₂ + 2Pb(NO₃)₂ combined and should divide it into the correct ratio?

Sorry if this is basic stuff, im quite new to chemistry and I haven't been able to find any similar examples to this.

[Borek]

CHeck the page I have linked to. Then think: how many moles of PbO₂ will be produced from 1 mole of Pb₃O₄? That's all in stoichiometric coefficients.

[NeonNe]

I get that, what about the 2Pb(NO₃)₂ which is also a product of the reaction?

[Borek]

Exactly the same approach. Look at the stoichiometric coefficients.

[NeonNe]

So 1 mole ----> 3 mole + 1.5 mole?

[Borek]

So 1 mole ----> 3 mole + 1.5 mole?

Hm, where did you get 3 and 1.5 from? Just look at the reaction equation:

Pb₃O₄ + 4HNO₃ -> PbO₂ + 2Pb(NO₃)₂ + 2H₂O

1 mole ----> . + .

What are the stoichiometric coefficients?

[NeonNe]

If I knew that then I can only assume I wouldn't be here trying to learn. I think you overestimate my understanding.

[Borek]

Looks like you are missing most basic things... Stoichiometric coefficients are part of the balanced reaction equation:

1Pb₃O₄ + 4HNO₃ -> 1PbO₂ + 2Pb(NO₃)₂ + 2H₂O

Nothing more fancy than just reading these numbers and using them in the way described on the page I linked to much earlier.

[NeonNe]

I appreciate your help. As I stated, im new to this.

So my understanding is 1 mole --> 1 mole + 2 mole? If so I think I have that done out in my workings from last night, I just wasn't sure if it was correct or not.

[Borek]

So my understanding is 1 mole --> 1 mole + 2 mole?

That's correct.

[NeonNe]

Great, I think I have that done already.

Pb₃O₄ (3g) + 4HNO₃ ----> PbO₂ + 2Pb(NO₃)₂ + 2H₂O

0.004mol ---> 0.004 mol + 0.008 mol

239.2g/mol X 0.004 = 0.95g PbO2
331.2g/mol X 0.008 = 2.65g 2Pb(NO₃)₂

2.65g was halved - 1.325g 2Pb(NO₃)₂ taken for next reaction


3Pb(NO₃)₂ (1.325g) + 2NaCO₃ + 2NaOH ---> Pb(OH)2.2PbCO₃ + 6NaNO₃

0.004 mol ---> 0.004 mol (1:1)

508.4g/mol X 0.004 = 2.03g Pb(OH)2.2PbCO₃

Am I on the right track now?

[Borek]

Much better, although 3 g of Pb₃O₄ is not 0.004 mole. Well - it is, if you round the number, but you shouldn't use rounded number in calculations, you round only the reported results.

And in the second reaction stoichiometric ratio is not 1:1.

[NeonNe]

Thanks I will amend the first part.

However in the second reaction there are 3 atoms of Pb on each side of the reaction, I don't get how the stoichiometric ratio could be different?