calculate the number of molecules of co2 produced when 10g of caco3 is treated with 100cm3 of 0.20mol dm cube hcl?

Robert01

5 years ago

2HCl + CaCO3 → CaCl2 + H2O + CO2↑

mass of limestone= 10g
no of moles of limestone= (10/100)mole= 0.1mol.
volume of HCl= (100/1000)dm³= 0.1dm³
molarity of HCl= 0.20mol/dm³
no of moles of HCl= (0.2 × 0.1) mol.= 0.02mol.

But 2 moles of HCl react with 1 mole of CaCO3
0.02 moles of HCl will react with x mole of CaCO3
x= 0.02 ÷ 2= 0.01mole.

And there are 0.1mole of CaCO3, thus, HCl got used up and it is the limiting reagent.

2 moles of HCl → 1 mole of CO2
0.02 mole of HCl → y
y= 0.02 ÷ 2= 0.01mol.

1 mole of CO2 → 6.02 × 10^23 molecules
0.02 mole of CO2 → z
z= 0.02 × 6.02 × 10^23= 1.204 × 10^22mole.