Limiting reactant problem
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Limiting reactant problem
What masses od CaCO3 and 20% HCl(aq) are needed to get 50,00 grams of calcium chloride hexahydrate? Can someone explain please?
- ChenBeier
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Re: Limiting reactant problem
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.
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.
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Re: Limiting reactant problem
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).
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
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Re: Limiting reactant problem
Generally your calculation path is correct but you mixed up some numbers.
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
Water is missing in the equationCaCO3 + 2HCl(aq) → CaCl2·6H2O + CO2(g)
Opposit 1 mol CaCO3 is equal 1 mol CaCl2. × 6 H2OAccording 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.
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
Re: Limiting reactant problem
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.
Re: Limiting reactant problem
Can someone explain to me more carefully when calculating the number of moles of CaCO3 and the mass of CaCO3?