The balanced equation for the above reaction is as follows;
CaCO₃ + 2HCl ----> CaCl₂ + H₂O + CO₂
stoichiometry of CaCO₃ to HCl is 1:2
molar volume states that 1 mol of any gas occupies a volume of 22.4 L at STP.
volume of 22.4 L occupied by 1 mol
therefore 0.56 L occupied by - 0.56 L / 22.4 L/mol = 0.025 mol
number of HCl moles reacted - 0.025 mol
2 mol of HCl reacts with 1 mol of CaCO₃
therefore 0.025 mol reacts with - 0.025/2 = 0.0125 mol
mass of CaCO₃ required - 0.0125 mol x 100 g/mol = 1.25 g
1.25 g of CaCO₃ is required
Answer: pH = 2.68
Explanation: pH = - log[H3O^+] .
HCl reacts with water : HCl + H2O -> Cl^- + H3O^+
So [H3O^+] = c(HCl) = 0.0021 M
pH = - log(0.0021) = 2.68
Answer:
Explanation:
It is given that,
Number of moles, n = 0.15
We need to find the number of particles in 0.15 mol of NaCl. Let N are the number of particles i.e.
Number of particles = number of moles × Avagadro's number
So,
So, the number of particles are .