Limiting reactant problem

[katjazoo]

What masses od CaCO₃ and 20% HCl(aq) are needed to get 50,00 grams of calcium chloride hexahydrate? Can someone explain please?

[ChenBeier]

First write chemical equation

Then convert 50g CaCl₂ x 6 H₂O into mol.
Read how many mol CaCO₃ correspond to it and calculate back to the mass. You need molare mass for it.
The same you do with HCl.
The mass of HCl you have to combine with percentage. Look up the specific gravity to get the volume.

[michaelnordquist]

Is my answer true or false?

To calculate the masses of CaCO₃ and 20% HCl(aq) needed to get 50.00 grams of calcium chloride hexahydrate, we need to use stoichiometry. The balanced chemical equation for the reaction is:

CaCO₃ + 2HCl(aq) → CaCl₂·6H₂O + CO₂(g)

The molar mass of CaCl₂·6H₂O is 219.08 g/mol. Therefore, 50.00 grams of CaCl₂·6H₂O is equal to 0.228 moles.

According to the balanced chemical equation, 1 mole of CaCO₃ reacts with 2 moles of HCl(aq) to produce 1 mole of CaCl₂·6H₂O. Therefore, we need 0.114 moles of CaCO₃ and 0.228 moles of HCl(aq) to produce 50.00 grams of CaCl₂·6H₂O.

The molar mass of CaCO₃ is 100.09 g/mol. Therefore, the mass of CaCO₃ needed is:

mass of CaCO₃ = 0.114 moles × 100.09 g/mol = 11.41 grams

To calculate the mass of 20% HCl(aq) needed, we need to first calculate the molarity of the solution. A 20% solution of HCl(aq) means that it contains 20 grams of HCl per 100 mL of solution. The density of the solution is 1.10 g/mL. Therefore, the molarity of the solution is:

molarity of HCl(aq) = (20 grams/36.46 g/mol) / (100 mL/1000 mL) = 0.547 M

The molar mass of HCl is 36.46 g/mol. Therefore, the mass of HCl(aq) needed is:

mass of HCl(aq) = 0.228 moles × 36.46 g/mol = 8.31 grams

Therefore, to get 50.00 grams of calcium chloride hexahydrate, we need 11.41 grams of CaCO₃ and 8.31 grams of 20% HCl(aq).

[ChenBeier]

Generally your calculation path is correct but you mixed up some numbers.

CaCO₃ + 2HCl(aq) → CaCl₂·6H₂O + CO₂(g)

Water is missing in the equation

According to the balanced chemical equation, 1 mole of CaCO₃ reacts with 2 moles of HCl(aq) to produce 1 mole of CaCl₂·6H₂O. Therefore, we need 0.114 moles of CaCO₃ and 0.228 moles of HCl(aq) to produce 50.00 grams of CaCl₂·6H₂O.

Opposit 1 mol CaCO₃ is equal 1 mol CaCl₂. × 6 H₂O
And 1 mol CaCO₃ is equal 2 HCl

The calculation of the 20% is also wrong.
20% means weight percentage. It means 20% = 20g/100g
With specific gravity the 100 g has to converted to ml. Means 100g/ 1,1 g/ ml = 90,91 ml
The 20 g has to converted to mol and this is the amount in 90,91 ml.
This express in mol/l

[farare]

What masses od CaCO₃ and 20% HCl(aq) are needed to get 50,00 grams of calcium chloride hexahydrate? Can someone explain please?

The limiting reactant problem is a crucial concept in chemistry that plays a fundamental role in determining the outcome of a chemical reaction. When multiple reactants are involved, the limiting reactant is the one that is completely consumed, thus limiting the amount of product that can be formed.

[Gomez22]

Can someone explain to me more carefully when calculating the number of moles of CaCO₃ and the mass of CaCO₃?