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# amount of a substance watch

1. An unknown metal carbonate reacts with a hydrochloric acid according to the following equation:
M2CO3(aq) +2HCl(aq) -----> 2MCl(aq) + CO2(g) + H2O(l)
A3.44 g sample of M2CO3was dissolved in distilled water to make 250 cm^3 of solution. A 25.0 cm^3 portion of this required 33.2 cm^3 of 0.150 mol dm^-3 hydrochloric acid for complete reaction.
part 1) calculate the amount, in moles, of HCl in 33.2 cm^3 of 0.150 mol dm^-3 hydrochloric acid.
part 2) calculate the amount, in moles of M2CO3 that reacted with this amount of HCl.
part 3) calculate the amount, in moles, of M2CO3 in the 3.4 g sample.
2. I presume you would do something along the lines:

1) Calculate moles of HCl reacted in 25cm^3 portion:
33.2 cm^3 = 0.0332 dm^3

0.0332 x 0.150 = 4.98 x 10^-3 moles

2) Calculate the scale factor in which the amount of solution has increased by:
33.2/25 = 1.328

3) Calculate new moles in 33.2 cm^3 solution:
1.328 x 4.98 x 10^-3 = 6.61 x 10^-3

This is what I would do if I saw this question but not 100% sure if this is correct.
3. (Original post by S.H.Rahman)
I presume you would do something along the lines:

1) Calculate moles of HCl reacted in 25cm^3 portion:
33.2 cm^3 = 0.0332 dm^3

0.0332 x 0.150 = 4.98 x 10^-3 moles

2) Calculate how scale factor in which the amount of solution has increased by:
33.2/25 = 1.328

3) Calculate new moles in 33.2 cm^3 solution:
1.328 x 4.98 x 10^-3 = 6.61 x 10^-3

This is what I would do if I saw this question but not 100% sure if this is correct.
thank you
4. (Original post by tats bentata)
An unknown metal carbonate reacts with a hydrochloric acid according to the following equation:
M2CO3(aq) +2HCl(aq) -----> 2MCl(aq) + CO2(g) + H2O(l)
A3.44 g sample of M2CO3was dissolved in distilled water to make 250 cm^3 of solution. A 25.0 cm^3 portion of this required 33.2 cm^3 of 0.150 mol dm^-3 hydrochloric acid for complete reaction.
part 1) calculate the amount, in moles, of HCl in 33.2 cm^3 of 0.150 mol dm^-3 hydrochloric acid.
part 2) calculate the amount, in moles of M2CO3 that reacted with this amount of HCl.
part 3) calculate the amount, in moles, of M2CO3 in the 3.4 g sample.
(Original post by S.H.Rahman)
I presume you would do something along the lines:

1) Calculate moles of HCl reacted in 25cm^3 portion:
33.2 cm^3 = 0.0332 dm^3

0.0332 x 0.150 = 4.98 x 10^-3 moles

2) Calculate how scale factor in which the amount of solution has increased by:
33.2/25 = 1.328

3) Calculate new moles in 33.2 cm^3 solution:
1.328 x 4.98 x 10^-3 = 6.61 x 10^-3

This is what I would do if I saw this question but not 100% sure if this is correct.

Answer to part 1 correct

Part 2 - the number of moles M2CO3 that reacted is (4.98 x 10^-3)/2 = 2.49 x 10^-3
This is because McCO3 reacts with HCl in a 1:2 ratio, this is from the equation.

Part 3 - you have the number of moles in 25cm3. The 3.4g sample was dissolved in 250cm3. Therefore there was 10x as many moles in the 3.4 g

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