Answer:
Mass of CaCl₂ produced = 15 g
Excess reactant = CaCO₃
Mass of CaCO₃ left = 11.5 g
Explanation:
Given data:
Mass of calcium carbonate = 25 g
Mass of hydrochloric acid = 10.0 g
Mass of calcium chloride produced = ?
Chemical equation:
CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
Number of moles of CaCO₃:
Number of moles of CaCO₃ = Mass /molar mass
Number of moles of CaCO₃= 25.0 g / 100.1 g/mol
Number of moles of CaCO₃ = 0.25 mol
Number of moles of HCl:
Number of moles of HCl = Mass /molar mass
Number of moles of HCl = 10.0 g / 36.5 g/mol
Number of moles of HCl = 0.27 mol
Now we will compare the moles of CaCl₂ with HCl and CaCO₃ .
CaCO₃ : CaCl₂
1 : 1
0.25 : 0.25
HCl : CaCl₂
2 : 1
0.27 : 1/2 × 0.27 = 0.135 mol
The number of moles of CaCl₂ produced by HCl are less it will be limiting reactant.
Mass of CaCl₂ = moles × molar mass
Mass of CaCl₂ =0.135 mol × 110.98 g/mol
Mass of CaCl₂ = 15 g
The calcium carbonate is present in excess.
HCl : CaCO₃
2 : 1
0.27 : 1/2 × 0.27 = 0.135 mol
So, 0.135 moles react with 0.27 moles of HCl.
The moles of CaCO₃ remain unreacted = 0.25 - 0.135
The moles of CaCO₃ remain unreacted = 0.115 mol
Mass remain unreacted:
Mass of of CaCO₃ remain unreacted = Moles × molar mass
Mass of of CaCO₃ remain unreacted = 0.115 mol × 100.1 g/mol
Mass of of CaCO₃ remain unreacted = 11.5 g
