calculate the number of mole of cacl that can be obtained from 30g of caco3 in the presence of excess Hcl(Ca =40,C=12,O=16,H=1,Cl=355)
You mean Cl = 35.5 g/mol but we do not need it
CaCO3 = 40 +12 + 3*16 = 100 g/mol
so 30/100 = 0.30 mol of CaCO3
same number of mols of CaCl because Ca has to be the same number of atoms left and right
so you get 0.30 mol of CaCl
To calculate the number of moles of CaCl2 that can be obtained from 30g of CaCO3 in the presence of excess HCl, we need to follow these steps:
Step 1: Convert the mass of CaCO3 to moles.
Molar mass of CaCO3:
Ca = 40 g/mol
C = 12 g/mol
O = 16 g/mol
Total molar mass of CaCO3 = (40 + 12 + (3 x 16)) g/mol = 100 g/mol
Number of moles of CaCO3 = mass of CaCO3 / molar mass of CaCO3
= 30 g / 100 g/mol
= 0.3 mol
Step 2: Determine the stoichiometry between CaCO3 and CaCl2.
The balanced chemical equation for the reaction between CaCO3 and HCl is:
CaCO3 + 2HCl → CaCl2 + H2O + CO2
From the equation, we can see that 1 mole of CaCO3 produces 1 mole of CaCl2.
Step 3: Calculate the number of moles of CaCl2 produced.
Number of moles of CaCl2 = number of moles of CaCO3
= 0.3 mol
Therefore, the number of moles of CaCl2 that can be obtained from 30g of CaCO3 in the presence of excess HCl is 0.3 mol.
To calculate the number of moles of CaCl2 that can be obtained from 30g of CaCO3, we need to follow these steps:
Step 1: Determine the molar mass of CaCO3.
The molar mass of CaCO3 can be calculated by adding up the atomic masses of the elements in its formula.
Molar mass of CaCO3 = (1 * Atomic mass of C) + (3 * Atomic mass of O) + (1 * Atomic mass of Ca)
= (1 * 12) + (3 * 16) + (1 * 40)
= 12 + 48 + 40
= 100 g/mol
Step 2: Calculate the number of moles of CaCO3 using its molar mass and given mass.
Number of moles of CaCO3 = Mass of CaCO3 / Molar mass of CaCO3
= 30g / 100 g/mol
= 0.3 mol
Step 3: Identify the stoichiometric ratio between CaCO3 and CaCl2 from the balanced chemical equation.
The balanced chemical equation is:
CaCO3 + 2HCl -> CaCl2 + H2O + CO2
From the equation, we can see that 1 mole of CaCO3 reacts with 1 mole of CaCl2.
Therefore, the number of moles of CaCl2 that can be obtained is the same as the number of moles of CaCO3, which is 0.3 mol.
Hence, the number of moles of CaCl2 that can be obtained from 30g of CaCO3 in the presence of excess HCl is 0.3 mol.
You probably know better and you're just lazy but you really need to learn to post/type better questions. That's HCl, CaCl2, CaCO3, Cl is 35.5 etc.
CaCO3 + 2HCl --> CaCl2 + H2O + CO2
mols CaCO3 = grams/molar mass = 30.0/100 = 0.300 mol
From the equation you can see we get 1 mol CaCl2 for every 1 mol CaCO3; therefore, we should expect 0.3 mol CaCl2 from 0.3 mol CaCO3.