5.0g of calcium carbonate were allowed to react with 25cm3 of 1.0M hydrochloric acid
Until there was no further reaction. Calculate the mass of calcium carbonate that
Remained Unreacted. (Ca =40.0, O=16.0, C= 12.0)
Freddy. All of this looks like a homework dump to me. You walk away and we do your work for you. After this please note that you need to show what you've tried and put forth SOME effort anyway. The freebie train just stopped. This is done the same way as the 17.0 g Zn problem. I'll be glad to check your work.
In order to calculate the mass of calcium carbonate that remained unreacted, we need to determine the limiting reactant in the reaction. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
To find the limiting reactant, we will compare the moles of calcium carbonate and hydrochloric acid.
First, let's calculate the number of moles of calcium carbonate:
Molar mass of CaCO3 = (40.0 g/mol) + (12.0 g/mol) + (3 * 16.0 g/mol) = 100.0 g/mol
Number of moles of CaCO3 = mass / molar mass = 5.0 g / 100.0 g/mol = 0.05 mol
Next, let's calculate the number of moles of hydrochloric acid:
Volume of HCl = 25.0 cm³ = 25.0 mL
Molarity of HCl = 1.0 M
Number of moles of HCl = volume (in liters) x molarity = 25.0 mL / 1000 mL/L x 1.0 mol/L = 0.025 mol
The balanced chemical equation for the reaction between calcium carbonate and hydrochloric acid is:
CaCO3 + 2HCl --> CaCl2 + H2O + CO2
According to the balanced equation, the ratio of calcium carbonate to hydrochloric acid is 1:2. This means that for every 1 mole of calcium carbonate, 2 moles of hydrochloric acid react.
Since the ratio is 1:2, we can see that 2 moles of hydrochloric acid are required to fully react with 1 mole of calcium carbonate.
Therefore, we can determine that the moles of calcium carbonate is the limiting reactant since we only have 0.05 mol, which is less than the 0.025 mol of hydrochloric acid.
Now, let's calculate the moles of calcium carbonate that reacted:
Moles of calcium carbonate reacted = moles of hydrochloric acid x (1 mole CaCO3 / 2 moles HCl)
= 0.025 mol x (1 mol CaCO3 / 2 mol HCl)
= 0.0125 mol
To find the moles of calcium carbonate that remained unreacted, we subtract the moles of calcium carbonate that reacted from the initial moles of calcium carbonate:
Moles of calcium carbonate unreacted = initial moles of calcium carbonate - moles of calcium carbonate reacted
= 0.05 mol - 0.0125 mol
= 0.0375 mol
Finally, we can calculate the mass of calcium carbonate that remained unreacted:
Mass of calcium carbonate unreacted = moles of calcium carbonate unreacted x molar mass
= 0.0375 mol x 100.0 g/mol
= 3.75 g
Therefore, the mass of calcium carbonate that remained unreacted is 3.75 grams.
To calculate the mass of calcium carbonate that remained unreacted, we need to determine the limiting reactant in the reaction and the amount of excess reactant.
Here are the steps to solve the problem:
1. Write the balanced chemical equation for the reaction between calcium carbonate (CaCO3) and hydrochloric acid (HCl):
CaCO3 + 2HCl → CaCl2 + H2O + CO2
2. Calculate the number of moles of calcium carbonate:
Molar mass of CaCO3 = (40.0 g/mol) + (12.0 g/mol) + (3 * 16.0 g/mol) = 100.0 g/mol
Number of moles of CaCO3 = mass of CaCO3 / molar mass of CaCO3 = 5.0 g / 100.0 g/mol = 0.05 mol
3. Calculate the number of moles of hydrochloric acid:
Number of moles of HCl = concentration of HCl * volume of HCl
Number of moles of HCl = 1.0 mol/L * (25 cm3 / 1000 cm3/L) = 0.025 mol
4. Determine the stoichiometric ratio between CaCO3 and HCl from the balanced equation:
From the balanced equation, the stoichiometric ratio of CaCO3 to HCl is 1:2. This means that 1 mole of CaCO3 reacts with 2 moles of HCl.
5. Determine the limiting reactant:
The limiting reactant is the reactant that will be completely consumed first and determines the amount of product formed. To find the limiting reactant, we need to compare the moles of CaCO3 and HCl.
Since the stoichiometric ratio is 1:2, for every mole of CaCO3, we need 2 moles of HCl. However, we only have 0.025 moles of HCl, which is less than the 0.05 moles of CaCO3. This means that HCl is the limiting reactant.
6. Calculate the amount of unreacted CaCO3:
Since HCl is the limiting reactant, it completely reacts, and the remaining CaCO3 is in excess.
The moles of CaCO3 remaining unreacted is calculated using the stoichiometric ratio of CaCO3 to HCl: 1 mole of CaCO3 reacts with 2 moles of HCl.
Number of moles of CaCO3 remaining unreacted = 0.05 mol (initial moles of CaCO3) - (2 moles of HCl consumed) = 0.05 mol - 0.025 mol = 0.025 mol
7. Calculate the mass of CaCO3 remaining unreacted:
Mass of CaCO3 remaining unreacted = number of moles of CaCO3 remaining unreacted * molar mass of CaCO3
Mass of CaCO3 remaining unreacted = 0.025 mol * 100.0 g/mol = 2.5 g
Therefore, the mass of calcium carbonate that remained unreacted is 2.5 grams.