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
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.
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.
Thank you for correcting the error in my response. Your solution is correct, and I apologize for any confusion caused by my mistake.
To find the relative molecular mass of the metal carbonate (Pcarbonate) and identify the metal (P), we need to follow a series of steps.
Step 1: Calculate the number of moles of HCl used.
Given: Volume of HCl solution = 100 cm^3
Concentration of HCl solution = 0.1 M
To find the number of moles of HCl, we use the formula:
Moles = Concentration × Volume
Moles of HCl = 0.1 M × (100 cm^3 / 1000)
Moles of HCl = 0.01 moles
Step 2: Calculate the number of moles of NaOH used.
Given: Volume of NaOH solution = 20 cm^3
Concentration of NaOH solution = 0.15 M
To find the number of moles of NaOH, we use the same formula:
Moles = Concentration × Volume
Moles of NaOH = 0.15 M × (20 cm^3 / 1000)
Moles of NaOH = 0.003 moles
Step 3: Determine the stoichiometry between HCl and NaOH.
From the balanced equation:
HCl + NaOH → NaCl + H2O
The stoichiometric ratio between HCl and NaOH is 1:1. This means that 1 mole of HCl reacts with 1 mole of NaOH.
Step 4: Determine the number of moles of HCl used in excess.
Since the number of moles of NaOH used is 0.003 moles and the stoichiometric ratio between HCl and NaOH is 1:1, the number of moles of HCl used in excess is also 0.003 moles.
Step 5: Calculate the number of moles of HCl that reacted with Pcarbonate.
The excess moles of HCl used in Step 4 is the difference between the total moles of HCl used (0.01 moles) and the moles of HCl that reacted with Pcarbonate.
Moles of HCl reacted with Pcarbonate = Total moles of HCl used - Excess moles of HCl used
Moles of HCl reacted with Pcarbonate = 0.01 moles - 0.003 moles
Moles of HCl reacted with Pcarbonate = 0.007 moles
Step 6: Determine the stoichiometry between HCl and Pcarbonate.
From the balanced equation:
2HCl + MCO3 → MCl2 + CO2 + H2O
The stoichiometric ratio between HCl and Pcarbonate is 2:1. This means that 2 moles of HCl react with 1 mole of Pcarbonate.
Step 7: Calculate the number of moles of Pcarbonate.
Using the stoichiometric ratio from Step 6, we can calculate the number of moles of Pcarbonate:
Moles of Pcarbonate = (Moles of HCl reacted with Pcarbonate) / 2
Moles of Pcarbonate = 0.007 moles / 2
Moles of Pcarbonate = 0.0035 moles
Step 8: Calculate the relative molecular mass of Pcarbonate.
The relative molecular mass (Mr) can be calculated using the formula:
Mr = (Mass of compound) / (Moles of compound)
Given that exactly 0.35g of Pcarbonate was dissolved, we can calculate the relative molecular mass:
Mr = 0.35g / 0.0035 moles
Mr = 100 g/mol
Therefore, the relative molecular mass of the metal carbonate Pcarbonate is 100 g/mol.
To identify the metal P, we need to know the formula of the metal carbonate. Based on the stoichiometry from Step 6, we know that 2 moles of HCl react with 1 mole of metal carbonate. Since 0.007 moles of HCl reacted with Pcarbonate, the formula of Pcarbonate is MCl2, where M represents the metal P.
To find the relative molecular mass of the metal carbonate (Pcarbonate) and identify the metal (P), we need to calculate the number of moles of the metal carbonate and use stoichiometry.
1. Calculate the number of moles of HCl used:
Moles of HCl = volume (L) x concentration (mol/L)
Moles of HCl = 0.1 mol/L x 0.100 L = 0.010 mol
2. Calculate the number of moles of NaOH used:
Moles of NaOH = volume (L) x concentration (mol/L)
Moles of NaOH = 0.15 mol/L x 0.020 L = 0.003 mol
3. Since the reaction between HCl and NaOH is 1:1, the moles of HCl used are equal to the moles of NaOH used for complete neutralization.
4. Calculate the number of moles of metal carbonate (Pcarbonate):
Moles of Pcarbonate = Moles of NaOH = 0.003 mol
5. Calculate the mass of Pcarbonate used:
Mass of Pcarbonate = moles x molar mass
Mass of Pcarbonate = 0.003 mol x molar mass (g/mol)
6. We are given that 0.35g of Pcarbonate was dissolved completely, so we can set up an equation:
0.003 mol x molar mass (g/mol) = 0.35g
7. Solve the equation for the molar mass:
Molar mass (g/mol) = 0.35g / 0.003 mol
8. Calculate the molar mass of Pcarbonate to find the relative molecular mass of the metal:
Molar mass (g/mol) = 116.67 g/mol
Therefore, the relative molecular mass of the metal carbonate (Pcarbonate) is approximately 116.67 g/mol. Since it is a divalent metal, the metal symbol P corresponds to "Calcium (Ca)" as it has a relative atomic mass close to 40.08 g/mol (half of 116.67 g/mol).