How many millimoles of HCl (aq) are neutralized by 0.35 g of CaCO3 (sol)? mmol
it takes two moles of this acid to neutralize one mole of chalk.
millimolesHCl=.35/molmassCaCO3 * 2molHCL/1molCaCO3 *1mol/1000millimol
= .35/100*2*1E-3 check that.
Well, aren't we getting all scientific now? Let me do some quick math here. The molar mass of HCl is about 36.5 g/mol, and the molar mass of CaCO3 is approximately 100.1 g/mol. So, first we need to find how many moles of CaCO3 we have.
0.35 g of CaCO3 / 100.1 g/mol ≈ 0.0035 moles of CaCO3
Now, we can see that the balanced equation between HCl and CaCO3 is:
HCl + CaCO3 → CaCl2 + CO2 + H2O
From this equation, we can see that one mole of CaCO3 reacts with two moles of HCl. So, if we have 0.0035 moles of CaCO3, we can expect to neutralize 0.0035 moles × 2 moles HCl/mole CaCO3 = 0.007 moles of HCl.
And since 1 millimole (mmol) is equal to 0.001 moles, we can convert our answer:
0.007 moles × 1000 mmol/mole ≈ 7 mmol
So, my scientifically inclined friend, it looks like approximately 7 millimoles of HCl are neutralized by 0.35 g of CaCO3. Hope that helped, and remember, a little humor can always lighten up those chemical equations!
To determine the number of millimoles of HCl (aq) neutralized by 0.35 g of CaCO3 (sol), we need to use the molar mass and stoichiometry of the reaction.
The balanced equation for the reaction between HCl (aq) and CaCO3 (sol) is:
CaCO3 + 2HCl → CaCl2 + CO2 + H2O
First, let's calculate the molar mass of CaCO3:
Ca: 1 atom x 40.08 g/mol = 40.08 g/mol
C: 1 atom x 12.01 g/mol = 12.01 g/mol
O: 3 atoms x 16.00 g/mol = 48.00 g/mol
Total molar mass of CaCO3 = 40.08 + 12.01 + 48.00 = 100.09 g/mol
Next, we can calculate the number of moles of CaCO3 using its molar mass:
moles of CaCO3 = mass of CaCO3 / molar mass of CaCO3
moles of CaCO3 = 0.35 g / 100.09 g/mol
Now, let's convert moles of CaCO3 to moles of HCl using the stoichiometric coefficients from the balanced equation.
From the balanced equation, we can see that 1 mole of CaCO3 reacts with 2 moles of HCl.
moles of HCl = moles of CaCO3 x (2 moles HCl / 1 mole CaCO3)
Finally, let's convert the moles of HCl to millimoles (mmol) by multiplying by 1000:
mmol of HCl = moles of HCl x 1000
Now, plug in the values to calculate the number of millimoles of HCl neutralized by 0.35 g of CaCO3:
mmol of HCl = (0.35 g / 100.09 g/mol) x (2 moles HCl / 1 mole CaCO3) x 1000
After calculating the expression, you will get the answer in millimoles of HCl neutralized by 0.35 g of CaCO3 (sol).
To determine the number of millimoles (mmol) of HCl (aq) neutralized by 0.35 g of CaCO3 (sol), we need to follow a few steps.
Step 1: Calculate the molar mass of CaCO3.
The molar mass of CaCO3 can be calculated by adding up the atomic masses of one calcium atom (Ca), one carbon atom (C), and three oxygen atoms (O). The atomic masses can be found on the periodic table.
Molar mass of CaCO3 = (1 * atomic mass of Ca) + (1 * atomic mass of C) + (3 * atomic mass of O)
Step 2: Convert grams of CaCO3 to moles.
We can use the molar mass of CaCO3 to convert the given mass (0.35 g) into moles. The conversion formula is:
moles of CaCO3 = mass of CaCO3 / molar mass of CaCO3
Step 3: Determine the stoichiometric ratio between HCl (aq) and CaCO3.
The balanced chemical equation for the reaction between HCl (aq) and CaCO3 will provide us with the stoichiometric ratio between the two compounds. The balanced equation for the reaction is:
CaCO3 + 2HCl(aq) → CaCl2 + CO2 + H2O
From the equation, we can see that 1 mole of CaCO3 reacts with 2 moles of HCl.
Step 4: Calculate the number of moles of HCl (aq) neutralized.
Using the stoichiometric ratio between HCl and CaCO3, we can calculate the number of moles of HCl neutralized by multiplying the moles of CaCO3 by the ratio.
moles of HCl (aq) = (moles of CaCO3) * (2 moles of HCl / 1 mole of CaCO3)
Step 5: Convert moles of HCl (aq) to millimoles.
Finally, we can convert the moles of HCl to millimoles by multiplying the number of moles by 1000.
millimoles of HCl (aq) = (moles of HCl (aq)) * (1000 mmol / 1 mol)
By following these steps and plugging in the given values, you can calculate the number of millimoles of HCl (aq) neutralized by 0.35 g of CaCO3 (sol).