How many milliliters of 0.237 M HCl are needed to react with 52.2g of CaCO3?
2HCl(aq) + CaCO3(s) --> CaCl2(aq) + CO2(g) + H2O(l)
mol wt of CaCO3 = 40.08 + 12.01 + 48 = 100.09
so, you have 52.2/100.09 = .522 moles
the equation indicates you need 2 moles of HCl for every mole of CaCO3, so you need 1.044 moles HCl
so, now you have to find out how many ml of .237M HCl contain 1.044 moles HCl
Obviously, 1000ml contain .237 moles. So, you need 1.044/.237 * 1000 = 4405.1 ml
Well, you've got quite the chemical equation there! Looks like there's some chemistry brewing. Now, let's figure out the amount of HCl needed to react with that CaCO3.
First, we need to find the number of moles of CaCO3. We can do that by dividing the mass (52.2g) by the molar mass of CaCO3. The molar mass of CaCO3 is 40.08 + (12.01 + 3(16.00)), which gives us 100.09 g/mol.
So, 52.2g ÷ 100.09 g/mol = 0.521 moles of CaCO3.
Now, let's go back to our chemical equation. We can see that the stoichiometric ratio between CaCO3 and HCl is 1:2. That means for every mole of CaCO3, we need 2 moles of HCl.
Therefore, we need 2 moles of HCl for every 0.521 moles of CaCO3.
To find the volume of HCl needed, we'll multiply the number of moles of HCl by its molarity (0.237 M). That gives us 0.521 moles of CaCO3 × 2 moles of HCl/1 mole of CaCO3 × 0.237 moles/L = 0.247 liters, or 247 milliliters.
So, approximately 247 milliliters of 0.237 M HCl are needed to react with 52.2g of CaCO3. Hope that was helpful!
To find the number of milliliters of 0.237 M HCl needed to react with 52.2 g of CaCO3, we need to use the stoichiometry of the balanced equation.
First, we need to convert the mass of CaCO3 to moles. The molar mass of CaCO3 is 100.09 g/mol.
moles of CaCO3 = mass of CaCO3 / molar mass of CaCO3
moles of CaCO3 = 52.2 g / 100.09 g/mol
moles of CaCO3 ≈ 0.521 mol
According to the balanced equation, the stoichiometric ratio between CaCO3 and HCl is 1:2 (from the coefficients of the balanced equation).
moles of HCl needed = moles of CaCO3 × 2
moles of HCl needed = 0.521 mol × 2
moles of HCl needed ≈ 1.042 mol
Now, we can calculate the volume of HCl needed using the molarity of 0.237 M HCl.
volume of HCl needed = moles of HCl needed / molarity of HCl
volume of HCl needed = 1.042 mol / 0.237 mol/L
volume of HCl needed ≈ 4.396 L
Finally, we need to convert the volume from liters to milliliters.
volume of HCl needed in milliliters = volume of HCl needed × 1000 mL/L
volume of HCl needed in milliliters ≈ 4.396 L × 1000 mL/L
volume of HCl needed in milliliters ≈ 4396 mL
Therefore, approximately 4396 milliliters of 0.237 M HCl are needed to react with 52.2 g of CaCO3.
To determine the number of milliliters of 0.237 M HCl needed to react with 52.2g of CaCO3, we'll first calculate the number of moles of CaCO3, and then use stoichiometry to find the amount of HCl required.
Step 1: Calculate the number of moles of CaCO3
The molar mass of CaCO3 is calculated by adding up the atomic masses of the elements:
Ca: 40.08 g/mol
C: 12.01 g/mol
O: 16.00 g/mol (there are 3 oxygen atoms)
Molar mass of CaCO3 = (40.08 g/mol) + (12.01 g/mol) + (16.00 g/mol × 3) = 100.09 g/mol
To find the number of moles of CaCO3, divide the given mass by its molar mass:
Number of moles of CaCO3 = 52.2 g / 100.09 g/mol = 0.5214 mol
Step 2: Use stoichiometry to find the number of moles of HCl
From the balanced equation, we can see that it takes 2 moles of HCl to react with 1 mole of CaCO3:
2 moles of HCl + 1 mole of CaCO3
Therefore, the number of moles of HCl needed is twice the number of moles of CaCO3:
Number of moles of HCl = 2 × (0.5214 mol) = 1.0428 mol
Step 3: Convert moles of HCl to milliliters using the given concentration
The concentration of HCl is 0.237 M, which means there are 0.237 moles of HCl in 1 liter (1000 milliliters) of solution.
To find the volume of HCl needed, divide the number of moles of HCl by its concentration:
Volume of HCl = Number of moles of HCl / Concentration of HCl
Volume of HCl = 1.0428 mol / 0.237 M = 4.402 mL
Therefore, you would need approximately 4.402 milliliters of 0.237 M HCl to react with 52.2 grams of CaCO3.