2HCl + M2CO3 >>> 2MCl + CO2 + H2O
So you reacted (25/1000)*7.46 grams of the carbonate, with .027*.1 moles of acid.
from the balanced equation, then the moles of M2CO3 is
1/2 * moles acid=1/2*.027*.1
figure that out.
Now you know (or can figure) the molmass of the carbonate
moles carbonate=above , then
moles carbonate=massused/molmass
or molmass=massused/moles carbonat
= 7.46g*(25/1000)/moles carbonate which you figured above.
Now haveing the mole mass of M2CO3, you can subtract the formula mass of CO3, and you are left with 2xAtomic mass M.
from that, determine the atomic mass of M.
A solution of a metal carbonate M2CO3, was prepared by dissolving 7.46 g of the anhydrous solid in water to give 1000cm^3 of solution. 25.0cm^3 of this solution reacted with 27.0cm^3 of 0.100 mol/dm^3 hydrochloric acid. Calculate the relative formula mass of M2CO3 and hence the relative atomic mass of the metal M
4 answers
mass M2CO3 titrated = 7.46 x (25/1000) = approx 0.19 but you need a more accurate number for this and the all of the calculations that follow.
M2CO3 + 2HCl ==> CO2 + H2O + 2MCl
mols HCl = M x L = approx 0.0027
mols M2CO3 = 1/2*0.0027 = 0.00135
mols = grams/molar mass or
molar mass = g/mol = approx 0.19/0.00135 = about 140.
2M = 140-C-3*O = 140-12-48 = about 80 for 2M or approx 40 for M. My guess is K at 39.1
M2CO3 + 2HCl ==> CO2 + H2O + 2MCl
mols HCl = M x L = approx 0.0027
mols M2CO3 = 1/2*0.0027 = 0.00135
mols = grams/molar mass or
molar mass = g/mol = approx 0.19/0.00135 = about 140.
2M = 140-C-3*O = 140-12-48 = about 80 for 2M or approx 40 for M. My guess is K at 39.1
I didn't understand the mass M2CO3 titrated = 7.46 x (25/1000) step. please, could you explain
You had 7.46 grams to begin. You placed that in 1000 cc and pulled out 25 cc of that. So the amount in the 25 cc is all you titrated (you didn't titrate the entire 7.46g sample) so how much sample was in that 25 cc. That's
7.46 x 25/1000 =
7.46 x 25/1000 =
3 answers