Some Help please?

Dee369369

Ok so iv my exam next week and can't do half the paper!
Could someone explain how to do this question.

What volume of gas is generated by the explosion of 10g of the black gunpowder?

C (s) + S(s) KNO3(s)

---

> CO2(g) + N2(g) + K2S(s)

I know that you add up the atomic weights but i was given the molar volume of gas and i don't know if i divide or multiply. and it doesn't fit any equations i know.

Also in triacylglycerols what does P, O, L and S stand for?

thanks for any help!;)

ZorbaTehZ

Well a bit of math on the calculator gave me 10g producing 6.66g of gas. If you balance the equation, and then add up the molecular weights of the charcoal and the nitrate and sulphur and then the gas on the other side, you should be able to find it.
Can't help you with the second part though.

Edit: Maybe this is what you're talking about? (unsaturated fatty acids)

P = palmitic; O = oleic; S = stearic; D = dodecanoic or lauric; C = capric; M = myristic; L = linoleic; Ln = linolenic; B = butyric.

resc

Dee369369 wrote: »

Ok so iv my exam next week and can't do half the paper!
Could someone explain how to do this question.

Hi

You need to work out how many moles of gas product form and then use the molar volume (22.4 L) to calculate the volume. Back of the envelope calculations (assuming KNO3 is the limiting reagent):

10 g KNO3 = 0.1 mol (assuming KNO3 molar mass = 103 g/mol)
Looking at balanced equation:2 moles of KNO3 give 3 moles of CO2 and 1 mole of N2 - ie 2 --> 4 this means 0.1 --> 0.2 mol

So the total number of moles of gas formed is 0.2 mol

Since 1 mole of gas = 22.4 L then 0.2 mol is 4.5 L.

Mucco

Dee369369 wrote: »

C (s) + S(s) KNO3(s)

---

> CO2(g) + N2(g) + K2S(s)

Balance the equation:
3C + S + 2KNO3

---

> 3CO2 + N2 + K2S

Calculate the combined molecular weight:
36 + 32 + 202 = 270 g/mol

Calculate no. moles of powder
10g/270g/mol = 37mmol

So 10g of black powder = 37 mmol of that combination
When this reacts, you get 4 equivs of gas
=148mmol
multiple by 22.4 l/mol to get volume (at standard pressure)

=3.3 litres

resc

Mucco wrote: »

Balance the equation:
3C + S + 2KNO3

---

> 3CO2 + N2 + K2S

Calculate the combined molecular weight:
36 + 32 + 202 = 270 g/mol

Calculate no. moles of powder
10g/270g/mol = 37mmol

So 10g of black powder = 37 mmol of that combination
When this reacts, you get 4 equivs of gas
=148mmol
multiple by 22.4 l/mol to get volume (at standard pressure)

=3.3 litres

Hi thanks, I was working off just KNO3, mistakenly. But 6 mol of reactant go to 4 mol of gas product, so is ratio not 6:4...?

ZorbaTehZ

Well I still think I'm right tbh - although I did give mass rather than volume :pac: The volume from my answer would be 5.53e-4 m^3.

resc

resc wrote: »

Hi thanks, I was working off just KNO3, mistakenly. But 6 mol of reactant go to 4 mol of gas product, so is ratio not 6:4...?

OK - I have double checked sums: Mucco is right. Have 10g on the reactant side, if this equates to 37 mmol, then we get 3 x 37 mmol of CO2 and 1 x 37 mmol of N2, 1 x 37 mmol of K2S. When you work out the mass of these you get 4.9g, 1 g and 4.1 g, giving 10g.

ZorbaTehZ

But why are you including the K2S since its a solid? I'm not trying to argue, it's just that I thought I knew how to do these questions properly (just had an exam in it)

resc

ZorbaTehZ wrote: »

But why are you including the K2S since its a solid? I'm not trying to argue, it's just that I thought I knew how to do these questions properly (just had an exam in it)

Hi

For the calculation you don't include the K2S, I was just showing that there was a mass balance (ie 10g goes to 10g). But for the calculation itself it is exactly how Mucco showed (although I am not too mad about the idea of grouping together all of the compounds and calling them "1 mole" - but I guess this is an easy trick to complete the question!)

Dee369369

O god lads ye've lost me! I tink i'll just concentrate on the other questions but thanks for helping!:)

Lucia1712

3?