First, we need to use stoichiometry to find the amount of HCl that reacted with the Pcarbonate:
0.1 mol/L HCl * 0.1 L = 0.01 mol HCl
Since Pcarbonate is divalent, it reacts with 2 moles of HCl:
0.01 mol HCl / 2 = 0.005 mol Pcarbonate
Next, we can use the molar mass of HCl to find the mass of Pcarbonate:
0.005 mol Pcarbonate * (2 mol Molar mass of HCl / 1 mol Pcarbonate) = 0.185 g Pcarbonate
Now, we can use the mass and concentration of the NaOH solution to find the amount of excess acid:
0.15 mol/L NaOH * 0.02 L = 0.003 mol NaOH
Since NaOH reacts with 1 mole of HCl:
0.003 mol NaOH = 0.003 mol HCl (excess)
The amount of HCl that reacted with Pcarbonate is:
0.01 mol HCl - 0.003 mol HCl = 0.007 mol HCl
Using the amount of Pcarbonate that we found earlier, we can find the molar mass:
0.185 g Pcarbonate / 0.005 mol Pcarbonate = 37 g/mol
Finally, we need to identify the metal P. Since Pcarbonate has the formula MCO3, we know that the metal has a 2+ charge. Looking at the periodic table, there are several possibilities for an element with a relative atomic mass of approximately 37, but two likely candidates are calcium (Ca) and strontium (Sr). Without more information, we cannot definitively identify the metal P.
Exactly 0.35g of divalent metal carbonate Pcarbonate were dissolved completely in 100 centimeter cubic of 0.1M HCL solution the excess acid required 20 centimeter cubic of 0.15M sodium hydroxide solution for complete neutralization . Find the relative molecular mass of the metal carbonate , identify metal P
3 answers
Bot has an error somewhere. Here is how you do it.
MCO3 + 2HCl ==> MCl2 + H2O + CO2
0.35 g + 0.1L x 0.1 M or
0.35 g + 0.01 mols HCl.
How much extra HCl was in the solution unused to dissolve MCO3? That's
HCl + NaOH = NaCl + H2O
0.01 - 0.15M x 0.02 L = 0.01 - 0.003 = 0.007 mols HCl used to dissolve the MCO3. Bot is OK to this point. To do the remainder,
How many moles MCO3 were there initially. That is
0.007 mols HCl x (1 mol MCO3/2 mols HCl) = 0.007/2 = 0.0035
mols MCO3 = grams MCO3/molar mass MCO3, then
mols MCO3 x molar mass MCO3 = grams MCO3 or
0.0035 x molar mass = 0.35 sample in the problem or
molar mass = 0.35/0.0035 = 100 = atomic mass M + C + 3*atomic mass O
molar mass MCO3 = 100 = M + 12 + 3*16
Thus atomic mass M = 40 which is Ca.
MCO3 + 2HCl ==> MCl2 + H2O + CO2
0.35 g + 0.1L x 0.1 M or
0.35 g + 0.01 mols HCl.
How much extra HCl was in the solution unused to dissolve MCO3? That's
HCl + NaOH = NaCl + H2O
0.01 - 0.15M x 0.02 L = 0.01 - 0.003 = 0.007 mols HCl used to dissolve the MCO3. Bot is OK to this point. To do the remainder,
How many moles MCO3 were there initially. That is
0.007 mols HCl x (1 mol MCO3/2 mols HCl) = 0.007/2 = 0.0035
mols MCO3 = grams MCO3/molar mass MCO3, then
mols MCO3 x molar mass MCO3 = grams MCO3 or
0.0035 x molar mass = 0.35 sample in the problem or
molar mass = 0.35/0.0035 = 100 = atomic mass M + C + 3*atomic mass O
molar mass MCO3 = 100 = M + 12 + 3*16
Thus atomic mass M = 40 which is Ca.
Thank you for correcting the error in my response. Your solution is correct, and I apologize for any confusion caused by my mistake.