The relative molecular mass of calcium carbonate is 100.
What is the minimum volume of 2.0 M hydrochloric acid that would be needed to completely react with 2.0 g of calcium carbonate?
options are
A) 20 cm^3
B) 10 cm^3
C) 5 cm^3
D) 30 cm^3
E) 40 cm^3
I wrote down what I believe is the balanced reaction: CaCO3 + 2HCl -> H2CO3 + CaCl2
but I'm not sure how to solve it
moles of CaCO3=0.02
by comparing moles of HCl=0.04
mass of Hcl=0.04*36.5=1.46g
we know that the molarity of Hcl solution is 2
and its percentage 7
SO
7g of HCl present in Hcl solution=100g
1g of HCl present in Hcl solution=100/7
1.46g of Hcl present in HCl solution=100/7*1.46=20.85
d=m/v
1.03g/cm3=20.85g/V
so it will be 20.24cm3
g HCl is OK but you don't need that. You know you need 0.04 mols HCl; therefore,
M HCl = mols HCl/L HCl
You know M and mols, solve for L.
That 7 stuff of percentage is hog wash.
To solve this problem, we need to calculate the amount of calcium carbonate that reacts with hydrochloric acid, and then determine the volume of hydrochloric acid required to react with that amount of calcium carbonate.
First, convert the mass of calcium carbonate (2.0 g) to moles by using its molar mass. The molar mass of calcium carbonate (CaCO3) is 40.08 g/mol for calcium (Ca), 12.01 g/mol for carbon (C), and 3(16.00 g/mol) for oxygen (O). Adding these together gives a molar mass of 100.08 g/mol for calcium carbonate.
Number of moles of calcium carbonate = mass / molar mass
Number of moles of calcium carbonate = 2.0 g / 100.08 g/mol
Now, we know that the balanced equation shows that 1 mole of calcium carbonate reacts with 2 moles of hydrochloric acid (HCl). Therefore, the number of moles of hydrochloric acid required will be twice the number of moles of calcium carbonate.
Number of moles of hydrochloric acid required = 2 x (number of moles of calcium carbonate)
Number of moles of hydrochloric acid required = 2 x (2.0 g / 100.08 g/mol)
Now, let's calculate the volume of hydrochloric acid required. We are given the concentration of hydrochloric acid as 2.0 M (moles per liter).
Volume of hydrochloric acid required (in liters) = number of moles of hydrochloric acid required / concentration of hydrochloric acid
Volume of hydrochloric acid required (in liters) = (2 x (2.0 g / 100.08 g/mol)) / 2.0 M
Finally, convert the volume from liters to cubic centimeters (cm^3) since that is the unit provided in the options.
Volume of hydrochloric acid required (in cm^3) = (volume of hydrochloric acid required in liters) x 1000 cm^3/L
Now, let's substitute the values and calculate the minimum volume of hydrochloric acid required:
Volume of hydrochloric acid required (in cm^3) = ((2 x (2.0 g / 100.08 g/mol)) / 2.0 M) x 1000 cm^3/L
Simplifying the expression, we get:
Volume of hydrochloric acid required (in cm^3) = (0.039976004 cm^3/M) x 1000 cm^3/L
Volume of hydrochloric acid required (in cm^3) ≈ 40 cm^3
Therefore, the minimum volume of 2.0 M hydrochloric acid needed to completely react with 2.0 g of calcium carbonate is approximately 40 cm^3.
The correct option from the given choices is E) 40 cm^3.
To determine the minimum volume of hydrochloric acid needed to react completely with calcium carbonate, we need to use the stoichiometry of the balanced equation.
The balanced equation you wrote is correct: CaCO3 + 2HCl -> H2CO3 + CaCl2
From the balanced equation, we can see that one mole of calcium carbonate reacts with two moles of hydrochloric acid.
First, we need to calculate the number of moles of calcium carbonate from the given mass:
mass of calcium carbonate = 2.0 g
molar mass of calcium carbonate = 100 g/mol (given)
number of moles of calcium carbonate = mass / molar mass = 2.0 g / 100 g/mol = 0.02 mol
Since the stoichiometry tells us that two moles of hydrochloric acid react with one mole of calcium carbonate, we know that 0.02 moles of calcium carbonate will react with 2 * 0.02 = 0.04 moles of hydrochloric acid.
Now, we can find the minimum volume of the 2.0 M (molar) hydrochloric acid using the formula:
Molarity (M) = moles / volume (L)
We can rearrange this equation to solve for volume:
volume (L) = moles / Molarity
Plugging in the values:
volume (L) = 0.04 mol / 2.0 mol/L = 0.02 L = 20 mL = 20 cm^3
Therefore, the minimum volume of 2.0 M hydrochloric acid needed to completely react with 2.0 g of calcium carbonate is 20 cm^3.
So, the correct option is A) 20 cm^3.