Help for chemistry

How would you go about working these out

It would be helpful to have the questions!

For number two, start by working out the number of moles of the quantity which you already know, and use stoichiometry to transfer that to the quantity which you do not know (in this case, the molar value is the same, and you can tell that because the equation does not have a number in front of any of the reactants (or products)). From there, it's just a case of re-arranging equations to work out what you don't know.

For number three, you're simply being asked to describe the process of titration, with the values given. You're making a standard volumetric solution, and this is described here.

Number four is similar to number two, just with slightly more complicated molecules - remember to use stoichiometry correctly, as you have a number in front of one of the reactants!

If you need me to be more specific (I've tried to give you general guidance without giving the answer away), let me know!

I'm still lost 😶

My notes are actually with my teacher at the moment, so I'm working solely off memory - forgive me if my explanation is not as clear as it could be!

For question two:

Firstly set out what we know already. We're reacting HCl with NH3, and it makes NH4Cl. All of these are in the aqueous state. We have 20 cm3 of ammonia solution, which is 0.02 dm3. That reacts with a certain volume of hydrochloric acid solution, which has a concentration of 0.02 mol dm-3. The hydrochloric acid which is titrated into the ammonia has a volume of 37.35 cm3, which is 0.03735 dm3 (which we'll round to 0.037 dm3). We can set this out like this to help us:

HCl (aq) + NH3 (aq) -> NH4Cl (aq)

HCl NH3
Vol. 0.037 0.01
Conc. 0.02

The unknown value which we now need to calculate is the concentration of NH3, and we can do that by working out the number of moles. The number of moles is equal to the concentration multiplied by the volume:

moles = 0.02 x 0.037 = 0.00074

Because there is only one molecule of each reactant in the equation, we can simply directly transfer this across, so there are also 0.00074 moles of NH3. Going back to our table:

HCl (aq) + NH3 (aq) -> NH4Cl (aq)

HCl NH3
Vol. 0.037 0.01
Conc. 0.02
Moles 0.00074 0.00074

We can now use that same formula: moles = concentration x volume, and re-arrange it so that we find out concentration. Concentration therefore equals moles divided by volume, so:

conc. = 0.00074 / 0.01 = 0.074 mol dm-3

This is our final answer. For the sake of completeness, we can fill it into our table:

HCl (aq) + NH3 (aq) -> NH4Cl (aq)

HCl NH3
Vol. 0.037 0.01
Conc. 0.02 0.074
Moles 0.00074 0.00074

For question three, have you read the process detailed in this document from AQA? It's pretty much exactly what you need.

For question four, you're basically doing the same thing as question two, but with different numbers and different molecules; hopefully my method detailed above will allow you to work this out for yourself!

EDIT: Unfortunately TSR's software removes excessive spaces, so my tables don't stick to their format. Sorry!

My notes are actually with my teacher at the moment, so I'm working solely off memory - forgive me if my explanation is not as clear as it could be!

For question two:

Firstly set out what we know already. We're reacting HCl with NH3, and it makes NH4Cl. All of these are in the aqueous state. We have 20 cm3 of ammonia solution, which is 0.02 dm3. That reacts with a certain volume of hydrochloric acid solution, which has a concentration of 0.02 mol dm-3. The hydrochloric acid which is titrated into the ammonia has a volume of 37.35 cm3, which is 0.03735 dm3 (which we'll round to 0.037 dm3). We can set this out like this to help us:

HCl (aq) + NH3 (aq) -> NH4Cl (aq)

HCl NH3
Vol. 0.037 0.01
Conc. 0.02

The unknown value which we now need to calculate is the concentration of NH3, and we can do that by working out the number of moles. The number of moles is equal to the concentration multiplied by the volume:

moles = 0.02 x 0.037 = 0.00074

Because there is only one molecule of each reactant in the equation, we can simply directly transfer this across, so there are also 0.00074 moles of NH3. Going back to our table:

HCl (aq) + NH3 (aq) -> NH4Cl (aq)

HCl NH3
Vol. 0.037 0.01
Conc. 0.02
Moles 0.00074 0.00074

We can now use that same formula: moles = concentration x volume, and re-arrange it so that we find out concentration. Concentration therefore equals moles divided by volume, so:

conc. = 0.00074 / 0.01 = 0.074 mol dm-3

This is our final answer. For the sake of completeness, we can fill it into our table:

HCl (aq) + NH3 (aq) -> NH4Cl (aq)

HCl NH3
Vol. 0.037 0.01
Conc. 0.02 0.074
Moles 0.00074 0.00074

For question three, have you read the process detailed in this document from AQA? It's pretty much exactly what you need.

For question four, you're basically doing the same thing as question two, but with different numbers and different molecules; hopefully my method detailed above will allow you to work this out for yourself!

EDIT: Unfortunately TSR's software removes excessive spaces, so my tables don't stick to their format. Sorry!

Could you work out number 4 quickly to see if I got the right answer or not please

In a spoiler so that anyone working through this in the future doesn't accidentally see the answer!

My notes are actually with my teacher at the moment, so I'm working solely off memory - forgive me if my explanation is not as clear as it could be!

For question two:

Firstly set out what we know already. We're reacting HCl with NH3, and it makes NH4Cl. All of these are in the aqueous state. We have 20 cm3 of ammonia solution, which is 0.02 dm3. That reacts with a certain volume of hydrochloric acid solution, which has a concentration of 0.02 mol dm-3. The hydrochloric acid which is titrated into the ammonia has a volume of 37.35 cm3, which is 0.03735 dm3 (which we'll round to 0.037 dm3). We can set this out like this to help us:

HCl (aq) + NH3 (aq) -> NH4Cl (aq)

HCl NH3
Vol. 0.037 0.01
Conc. 0.02

The unknown value which we now need to calculate is the concentration of NH3, and we can do that by working out the number of moles. The number of moles is equal to the concentration multiplied by the volume:

moles = 0.02 x 0.037 = 0.00074

Because there is only one molecule of each reactant in the equation, we can simply directly transfer this across, so there are also 0.00074 moles of NH3. Going back to our table:

HCl (aq) + NH3 (aq) -> NH4Cl (aq)

HCl NH3
Vol. 0.037 0.01
Conc. 0.02
Moles 0.00074 0.00074

We can now use that same formula: moles = concentration x volume, and re-arrange it so that we find out concentration. Concentration therefore equals moles divided by volume, so:

conc. = 0.00074 / 0.01 = 0.074 mol dm-3

This is our final answer. For the sake of completeness, we can fill it into our table:

HCl (aq) + NH3 (aq) -> NH4Cl (aq)

HCl NH3
Vol. 0.037 0.01
Conc. 0.02 0.074
Moles 0.00074 0.00074

For question three, have you read the process detailed in this document from AQA? It's pretty much exactly what you need.

For question four, you're basically doing the same thing as question two, but with different numbers and different molecules; hopefully my method detailed above will allow you to work this out for yourself!

EDIT: Unfortunately TSR's software removes excessive spaces, so my tables don't stick to their format. Sorry!

I thought we had 25 cm of solution and not 20?

Could you go through it but only showing the working out please

You're right, it was 25 cm3, not 20 cm3 - apparently I can't read - but the method you'd follow is exactly the same!



The method is exactly the same as for question two - what exactly don't you understand?

You're right, it was 25 cm3, not 20 cm3 - apparently I can't read - but the method you'd follow is exactly the same!



The method is exactly the same as for question two - what exactly don't you understand?

Also I redone the q2 and got 0.0925 mol Dm-3 is that right

Will I have to divide by 0.01 as well for q4?

Yes, that's correct.



The 0.01 comes from the calculation of the concentration; it's the volume of NH3. In question four, you're dividing by the volume of nitric acid (25 cm3, or 0.025 dm3).