Andiio's Chem Questions!

[Andiio]

0.012g of powdered magnesium was added to 250 mL of 0.010 M hydrochloric acid. (This was more than enough hydrochloric acid to react completely with all the magnesium)
b) What amount, in mole, of hydrochloric acid will have been used up when the reaction is complete?

I think I'm confusing myself by overthinking/overanalysing the question, but thanks!

[thushan]

We know that HCl (aq) is in excess.

This is a redox reaction, where H+ is reduced to H2 gas. Also, by balancing the equation we know that n(HCl)[reacted] = 2n(Mg).
So n(Mg) = 0.012/24.3 = 4.94 x 10^(-4) mol.
Hence n(HCl) = 9.9 x 10^(-4) mol.

Hence 9.9 x 10^04 mol of HCl would have been used up.

[Andiio]

We know that HCl (aq) is in excess.

This is a redox reaction, where H+ is reduced to H2 gas. Also, by balancing the equation we know that n(HCl)[reacted] = 2n(Mg).
So n(Mg) = 0.012/24.3 = 4.94 x 10^(-4) mol.
Hence n(HCl) = 9.9 x 10^(-4) mol.

Hence 9.9 x 10^04 mol of HCl would have been used up.

But when you balance the eq, isn't it:

2Mg (s) + 2HCl (aq) -> 2MgCl (aq) + H2(g)    ?

thus wouldn't n(HCl) = n(Mg) = 4.94 x 10^-4 mol?

[stonecold]

We know that HCl (aq) is in excess.

This is a redox reaction, where H+ is reduced to H2 gas. Also, by balancing the equation we know that n(HCl)[reacted] = 2n(Mg).
So n(Mg) = 0.012/24.3 = 4.94 x 10^(-4) mol.
Hence n(HCl) = 9.9 x 10^(-4) mol.

Hence 9.9 x 10^04 mol of HCl would have been used up.

But when you balance the eq, isn't it:

2Mg (s) + 2HCl (aq) -> 2MgCl (aq) + H2(g)    ?

thus wouldn't n(HCl) = n(Mg) = 4.94 x 10^-4 mol?

Nah.

It is this:

Mg (s) + 2HCl (aq) -> MgCl2 (aq) + H2(g)

[Andiio]

We know that HCl (aq) is in excess.

This is a redox reaction, where H+ is reduced to H2 gas. Also, by balancing the equation we know that n(HCl)[reacted] = 2n(Mg).
So n(Mg) = 0.012/24.3 = 4.94 x 10^(-4) mol.
Hence n(HCl) = 9.9 x 10^(-4) mol.

Hence 9.9 x 10^04 mol of HCl would have been used up.

But when you balance the eq, isn't it:

2Mg (s) + 2HCl (aq) -> 2MgCl (aq) + H2(g)    ?

thus wouldn't n(HCl) = n(Mg) = 4.94 x 10^-4 mol?

Nah.

It is this:

Mg (s) + 2HCl (aq) -> MgCl2 (aq) + H2(g)

OMG WOOPS FORGOT ABOUT VALENCIES LOL

[Andiio]

20.0 mL of 0.10 M hydrochloric acid (HCl) reacts with 20.0 mL of 0.30 M potassium hydroxide (KOH) solution.
The concentration of potassium ions in the resultant solution, in mole per litre, is?


I've written the equation - KOH + HCl -> KCl + H2O ; but mind-blanked and not sure what to do.

Thanks!

[pi]

Haven't done this stuff since last year, so am probably wrong:

KOH + HCl --> KCl + H2O
n(KOH) = cV = 0.30(0.0200) = 0.006 mol
V(final products) = 40.0 mL
c(K+) = n/V = 0.006/0.040 = 0.15 mol/L

(probably wrong though...) Was actually right. TIME TO START STUDYING CHEM DAMN IT!

[luken93]

n(HCl) = 0.10 x 0.020 = 0.002 mol
n(KOH) = 0.30 x 0.020 = 0.006 mol

According to the equation, the ratio of the reactants is 1:1, meaning that the KOH is in excess.

See if you can go from there...

[thushan]

No need to even consider the reaction, because K+ is a spectator ion and would not take part in the reaction.
So just find n(K+) by finding n(KOH), and from there you can find [K+].

[pi]

No need to even consider the reaction, because K+ is a spectator ion and would not take part in the reaction.
So just find n(K+) by finding n(KOH), and from there you can find [K+].

Does that mean I was on the right track?

I have forgotten all this stuff completely

[luken93]

No need to even consider the reaction, because K+ is a spectator ion and would not take part in the reaction.
So just find n(K+) by finding n(KOH), and from there you can find [K+].

Does that mean I was on the right track?

I have forgotten all this stuff completely

Yeah yours is correct, as you've found the n(K+) to start with / total volume = concentration in solution

Moderator action: removed real name, sorry for the inconvenience

[Andiio]

Thanks everyone!

Here's another question - just want to see how everyone here would approach it.



To determine the empirical formula of an ester (containing only carbon, hydrogen and oxygen), 1.02g of the ester was burnt completely in excess oxygen. The only products formed were 2.20 g of carbon dioxide and 0.90 g of water vapour.

Calculate the mass of carbon in 1.02g of the compound.



Thanks!

[samiira]

i though it would be :

mass of CO2 = 2.20g

mole of CO2 = mass/Mr = 2.2/44 = 0.050 moles
mole of Carbon = mole of CO2 = 0.050 moles

therefore mass of Carbon = 0.05 x 12 = 0.60g

i guess i cud be wrong ??

[man0005]

I think sammi's method is right
with luken's way, don't think you can assume that all the oxygen in the products will be present in the ester since it is burned in excess oxygen.
i could be wrong though

[david10d]

Shouldn't luken's method be correct as he has used the limiting reactant? That's how I would've worked it out.

I thought you couldn't assume that the mole ratio is correct without knowing the full equation and empirical formula of the ester (may or may not be correct).