Answer:
The mass percentage of CaCO3(s) in the sample is 83.4%
Explanation:
CaCO₃ + 2HCl → CaCl₂ + H2O + CO₂
Let's find out the moles of each reactant:
[HCl] = 0.150 mol/L
Molarity . volume = moles
0.150 m/L . 0.050L = 7.5x10⁻³ moles HCl
Molar mass of CaCO₃ =100.08 g
Moles of CaCO₃ = Mass / Molar mass
0.450 g/100.08 g = 4.49x10⁻³ moles of salt
Ratio is 1:2
If 2 moles of HCl react with 1 mol of salt
7.5x10⁻³ moles of HCl react with 7.5x10⁻³ / 2 = 3.75x10⁻³ moles
We must find out the mass
Moles of CaCO₃ . Molar mass CO₃ = 0.3753 g
So now, the mass percentage of salt will be
0.450g ___ 100%
0.3753 g ____ (0.3753 .100 )/0.450 =83.4 %
